Chapter 2 Linear Equations in One Variable

We are asked to solve the linear equation for the variable $x$.

The given equation is: $2x - 3 = x + 2$

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To solve for $x$, we need to gather the terms containing $x$ on one side of the equation and the constant terms on the other side.

Subtract $x$ from both sides of the equation to move the $x$ term from the right side to the left side:

$2x - 3 - x = x + 2 - x$

Simplifying gives: $x - 3 = 2$

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Now, add 3 to both sides of the equation to move the constant term from the left side to the right side and isolate $x$:

$x - 3 + 3 = 2 + 3$

Simplifying gives: $x = 5$

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Thus, the solution to the equation is $\mathbf{x = 5}$.

Verification: Substitute $x = 5$ into the original equation.

LHS $= 2(5) - 3 = 10 - 3 = 7$

RHS $= 5 + 2 = 7$

Since LHS $=$ RHS, the solution is correct.

We are asked to solve the linear equation for the variable $x$.

The given equation is: $5x + \frac{7}{2} = \frac{3}{2} x - 14$

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To make the equation easier to work with, we can remove the fractions by multiplying both sides by the Least Common Multiple (LCM) of the denominators, which is 2.

Multiply both sides of the equation by 2:

$2 \times \left( 5x + \frac{7}{2} \right) = 2 \times \left( \frac{3}{2} x - 14 \right)$

Applying the distributive property and simplifying:

$2 \times 5x + \cancel{2} \times \frac{7}{\cancel{2}} = \cancel{2} \times \frac{3}{\cancel{2}} x - 2 \times 14$

$10x + 7 = 3x - 28$

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Now, collect the terms with $x$ on one side and the constant terms on the other side. Subtract $3x$ from both sides:

$10x + 7 - 3x = 3x - 28 - 3x$

Simplifying gives: $7x + 7 = -28$

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Subtract 7 from both sides to isolate the $x$ term:

$7x + 7 - 7 = -28 - 7$

Simplifying gives: $7x = -35$

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Finally, divide both sides by 7 to solve for $x$:

$\frac{7x}{7} = \frac{-35}{7}$

$x = -5$

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Thus, the solution to the equation is $\mathbf{x = -5}$.

Verification: Substitute $x = -5$ into the original equation.

LHS $= 5(-5) + \frac{7}{2} = -25 + \frac{7}{2} = \frac{-50}{2} + \frac{7}{2} = \frac{-43}{2}$

RHS $= \frac{3}{2}(-5) - 14 = \frac{-15}{2} - 14 = \frac{-15}{2} - \frac{28}{2} = \frac{-43}{2}$

Since LHS $=$ RHS, the solution is correct.

Given:

The equation: $3x = 2x + 18$

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To Solve:

Find the value of $x$.

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Solution:

We are given the equation:

$3x = 2x + 18$

To isolate the variable $x$, we need to move the terms containing $x$ to one side of the equation.

Subtract $2x$ from both sides of the equation:

$3x - 2x = 2x + 18 - 2x$

Simplify both sides of the equation:

$1x = 18$

Which simplifies to:

$x = 18$

So, the solution to the equation is $x = 18$.

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Check:

To check the result, substitute $x = 18$ back into the original equation $3x = 2x + 18$.

Left Hand Side (LHS):

LHS $= 3x$

Substitute $x = 18$:

LHS $= 3 \times 18$

LHS $= 54$

Right Hand Side (RHS):

RHS $= 2x + 18$

Substitute $x = 18$:

RHS $= 2 \times 18 + 18$

RHS $= 36 + 18$

RHS $= 54$

Since LHS = RHS ($54 = 54$), the solution $x = 18$ is correct.

Given:

The equation: $5t - 3 = 3t - 5$

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To Solve:

Find the value of $t$.

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Solution:

We are given the equation:

$5t - 3 = 3t - 5$

To solve for $t$, we need to bring all terms involving $t$ to one side and all constant terms to the other side.

Subtract $3t$ from both sides of the equation:

$5t - 3 - 3t = 3t - 5 - 3t$

Simplify both sides:

$2t - 3 = -5$

Now, add $3$ to both sides of the equation:

$2t - 3 + 3 = -5 + 3$

Simplify both sides:

$2t = -2$

Now, divide both sides by $2$ to isolate $t$:

$\frac{2t}{2} = \frac{-2}{2}$

Simplify both sides:

$t = -1$

So, the solution to the equation is $t = -1$.

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Check:

To check the result, substitute $t = -1$ back into the original equation $5t - 3 = 3t - 5$.

Left Hand Side (LHS):

LHS $= 5t - 3$

Substitute $t = -1$:

LHS $= 5(-1) - 3$

LHS $= -5 - 3$

LHS $= -8$

Right Hand Side (RHS):

RHS $= 3t - 5$

Substitute $t = -1$:

RHS $= 3(-1) - 5$

RHS $= -3 - 5$

RHS $= -8$

Since LHS = RHS ($-8 = -8$), the solution $t = -1$ is correct.

Given:

The equation: $5x + 9 = 5 + 3x$

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To Solve:

Find the value of $x$.

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Solution:

We are given the equation:

$5x + 9 = 5 + 3x$

To solve for $x$, we will transpose the terms containing the variable $x$ to one side and the constant terms to the other side.

Subtract $3x$ from both sides of the equation:

$5x + 9 - 3x = 5 + 3x - 3x$

Simplify both sides:

$2x + 9 = 5$

Now, subtract $9$ from both sides of the equation:

$2x + 9 - 9 = 5 - 9$

Simplify both sides:

$2x = -4$

Now, divide both sides by $2$ to isolate $x$:

$\frac{2x}{2} = \frac{-4}{2}$

Simplify both sides:

$x = -2$

So, the solution to the equation is $x = -2$.

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Check:

To check the result, substitute $x = -2$ back into the original equation $5x + 9 = 5 + 3x$.

Left Hand Side (LHS):

LHS $= 5x + 9$

Substitute $x = -2$:

LHS $= 5(-2) + 9$

LHS $= -10 + 9$

LHS $= -1$

Right Hand Side (RHS):

RHS $= 5 + 3x$

Substitute $x = -2$:

RHS $= 5 + 3(-2)$

RHS $= 5 - 6$

RHS $= -1$

Since LHS = RHS ($-1 = -1$), the solution $x = -2$ is correct.

Given:

The equation: $4z + 3 = 6 + 2z$

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To Solve:

Find the value of $z$.

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Solution:

We are given the equation:

$4z + 3 = 6 + 2z$

To solve for $z$, we will move the terms containing the variable $z$ to one side and the constant terms to the other side.

Subtract $2z$ from both sides of the equation:

$4z + 3 - 2z = 6 + 2z - 2z$

Simplify both sides:

$2z + 3 = 6$

Now, subtract $3$ from both sides of the equation:

$2z + 3 - 3 = 6 - 3$

Simplify both sides:

$2z = 3$

Now, divide both sides by $2$ to isolate $z$:

$\frac{2z}{2} = \frac{3}{2}$

Simplify both sides:

$z = \frac{3}{2}$

So, the solution to the equation is $z = \frac{3}{2}$.

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Check:

To check the result, substitute $z = \frac{3}{2}$ back into the original equation $4z + 3 = 6 + 2z$.

Left Hand Side (LHS):

LHS $= 4z + 3$

Substitute $z = \frac{3}{2}$:

LHS $= 4\left(\frac{3}{2}\right) + 3$

LHS $= \frac{12}{2} + 3$

LHS $= 6 + 3$

LHS $= 9$

Right Hand Side (RHS):

RHS $= 6 + 2z$

Substitute $z = \frac{3}{2}$:

RHS $= 6 + 2\left(\frac{3}{2}\right)$

RHS $= 6 + \frac{6}{2}$

RHS $= 6 + 3$

RHS $= 9$

Since LHS = RHS ($9 = 9$), the solution $z = \frac{3}{2}$ is correct.

Given:

The equation: $2x - 1 = 14 - x$

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To Solve:

Find the value of $x$.

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Solution:

We are given the equation:

$2x - 1 = 14 - x$

To solve for $x$, we need to collect the terms with $x$ on one side and the constant terms on the other side.

Add $x$ to both sides of the equation:

$2x - 1 + x = 14 - x + x$

Simplify both sides:

$3x - 1 = 14$

Now, add $1$ to both sides of the equation:

$3x - 1 + 1 = 14 + 1$

Simplify both sides:

$3x = 15$

Now, divide both sides by $3$ to isolate $x$:

$\frac{3x}{3} = \frac{15}{3}$

Simplify both sides:

$x = 5$

So, the solution to the equation is $x = 5$.

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Check:

To check the result, substitute $x = 5$ back into the original equation $2x - 1 = 14 - x$.

Left Hand Side (LHS):

LHS $= 2x - 1$

Substitute $x = 5$:

LHS $= 2(5) - 1$

LHS $= 10 - 1$

LHS $= 9$

Right Hand Side (RHS):

RHS $= 14 - x$

Substitute $x = 5$:

RHS $= 14 - 5$

RHS $= 9$

Since LHS = RHS ($9 = 9$), the solution $x = 5$ is correct.

Given:

The equation: $8x + 4 = 3 (x - 1) + 7$

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To Solve:

Find the value of $x$.

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Solution:

We are given the equation:

$8x + 4 = 3 (x - 1) + 7$

First, simplify the right side of the equation by distributing the $3$:

$8x + 4 = 3x - 3 + 7$

Combine the constant terms on the right side:

$8x + 4 = 3x + 4$

Now, collect the terms with $x$ on one side and constant terms on the other.

Subtract $3x$ from both sides of the equation:

$8x + 4 - 3x = 3x + 4 - 3x$

Simplify both sides:

$5x + 4 = 4$

Now, subtract $4$ from both sides of the equation:

$5x + 4 - 4 = 4 - 4$

Simplify both sides:

$5x = 0$

Now, divide both sides by $5$ to isolate $x$:

$\frac{5x}{5} = \frac{0}{5}$

Simplify both sides:

$x = 0$

So, the solution to the equation is $x = 0$.

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Check:

To check the result, substitute $x = 0$ back into the original equation $8x + 4 = 3 (x - 1) + 7$.

Left Hand Side (LHS):

LHS $= 8x + 4$

Substitute $x = 0$:

LHS $= 8(0) + 4$

LHS $= 0 + 4$

LHS $= 4$

Right Hand Side (RHS):

RHS $= 3 (x - 1) + 7$

Substitute $x = 0$:

RHS $= 3 (0 - 1) + 7$

RHS $= 3 (-1) + 7$

RHS $= -3 + 7$

RHS $= 4$

Since LHS = RHS ($4 = 4$), the solution $x = 0$ is correct.

Given:

The equation: $x = \frac{4}{5} (x + 10)$

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To Solve:

Find the value of $x$.

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Solution:

We are given the equation:

$x = \frac{4}{5} (x + 10)$

To eliminate the fraction, multiply both sides of the equation by $5$:

$5 \times x = 5 \times \frac{4}{5} (x + 10)$

Simplify both sides:

$5x = 4(x + 10)$

Now, distribute the $4$ on the right side of the equation:

$5x = 4x + 4 \times 10$

Simplify the right side:

$5x = 4x + 40$

To solve for $x$, collect the terms containing $x$ on one side.

Subtract $4x$ from both sides of the equation:

$5x - 4x = 4x + 40 - 4x$

Simplify both sides:

$1x = 40$

Which simplifies to:

$x = 40$

So, the solution to the equation is $x = 40$.

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Check:

To check the result, substitute $x = 40$ back into the original equation $x = \frac{4}{5} (x + 10)$.

Left Hand Side (LHS):

LHS $= x$

Substitute $x = 40$:

LHS $= 40$

Right Hand Side (RHS):

RHS $= \frac{4}{5} (x + 10)$

Substitute $x = 40$:

RHS $= \frac{4}{5} (40 + 10)$

RHS $= \frac{4}{5} (50)$

Now, calculate the value:

RHS $= \frac{4}{\cancel{5}_1} \times \cancel{50}^{10}$

RHS $= 4 \times 10$

RHS $= 40$

Since LHS = RHS ($40 = 40$), the solution $x = 40$ is correct.

Given:

The equation: $\frac{2x}{3} + 1 = \frac{7x}{15} + 3$

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To Solve:

Find the value of $x$.

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Solution:

We are given the equation:

$\frac{2x}{3} + 1 = \frac{7x}{15} + 3$

To eliminate the denominators, we find the Least Common Multiple (LCM) of the denominators $3$ and $15$.

The multiples of $3$ are $3, 6, 9, 12, 15, ...$

The multiples of $15$ are $15, 30, 45, ...$

The LCM of $3$ and $15$ is $15$.

Multiply both sides of the equation by the LCM, $15$:

$\cancel{15}^{5} \times \frac{2x}{\cancel{3}_1} + 15 \times 1 = \cancel{15}^{1} \times \frac{7x}{\cancel{15}_1} + 15 \times 3$

(Multiplying both sides by 15)

Simplify both sides:

$5 \times 2x + 15 = 1 \times 7x + 45$

$10x + 15 = 7x + 45$

Now, collect the terms containing $x$ on one side and constant terms on the other.

Subtract $7x$ from both sides:

$10x + 15 - 7x = 7x + 45 - 7x$

Simplify both sides:

$3x + 15 = 45$

Now, subtract $15$ from both sides:

$3x + 15 - 15 = 45 - 15$

Simplify both sides:

$3x = 30$

Now, divide both sides by $3$ to isolate $x$:

$\frac{3x}{3} = \frac{30}{3}$

Simplify both sides:

$x = 10$

So, the solution to the equation is $x = 10$.

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Check:

To check the result, substitute $x = 10$ back into the original equation $\frac{2x}{3} + 1 = \frac{7x}{15} + 3$.

Left Hand Side (LHS):

LHS $= \frac{2x}{3} + 1$

Substitute $x = 10$:

LHS $= \frac{2(10)}{3} + 1$

LHS $= \frac{20}{3} + 1$

To add the fraction and the integer, find a common denominator:

LHS $= \frac{20}{3} + \frac{3}{3}$

LHS $= \frac{20 + 3}{3}$

LHS $= \frac{23}{3}$

Right Hand Side (RHS):

RHS $= \frac{7x}{15} + 3$

Substitute $x = 10$:

RHS $= \frac{7(10)}{15} + 3$

RHS $= \frac{70}{15} + 3$

Simplify the fraction $\frac{70}{15}$ by dividing the numerator and denominator by their greatest common divisor, which is $5$:

RHS $= \frac{\cancel{70}^{14}}{\cancel{15}_3} + 3$

RHS $= \frac{14}{3} + 3$

To add the fraction and the integer, find a common denominator:

RHS $= \frac{14}{3} + \frac{9}{3}$

RHS $= \frac{14 + 9}{3}$

RHS $= \frac{23}{3}$

Since LHS = RHS ($\frac{23}{3} = \frac{23}{3}$), the solution $x = 10$ is correct.

Given:

The equation: $2y + \frac{5}{3} = \frac{26}{3} - y$

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To Solve:

Find the value of $y$.

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Solution:

We are given the equation:

$2y + \frac{5}{3} = \frac{26}{3} - y$

To eliminate the fractions, we can multiply both sides of the equation by the Least Common Multiple (LCM) of the denominators, which is $3$.

$3 \times \left(2y + \frac{5}{3}\right) = 3 \times \left(\frac{26}{3} - y\right)$

(Multiplying both sides by 3)

Distribute the $3$ on both sides:

$3 \times 2y + 3 \times \frac{5}{3} = 3 \times \frac{26}{3} - 3 \times y$

(Distributing 3)

Simplify the equation:

$6y + 5 = 26 - 3y$

Now, collect the terms containing $y$ on one side and the constant terms on the other side.

Add $3y$ to both sides of the equation:

$6y + 5 + 3y = 26 - 3y + 3y$

Simplify both sides:

$9y + 5 = 26$

Now, subtract $5$ from both sides of the equation:

$9y + 5 - 5 = 26 - 5$

Simplify both sides:

$9y = 21$

Now, divide both sides by $9$ to isolate $y$:

$\frac{9y}{9} = \frac{21}{9}$

Simplify the fraction on the right side by dividing the numerator and denominator by their greatest common divisor, which is $3$:

$y = \frac{\cancel{21}^7}{\cancel{9}_3}$

So, the solution to the equation is $y = \frac{7}{3}$.

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Check:

To check the result, substitute $y = \frac{7}{3}$ back into the original equation $2y + \frac{5}{3} = \frac{26}{3} - y$.

Left Hand Side (LHS):

LHS $= 2y + \frac{5}{3}$

Substitute $y = \frac{7}{3}$:

LHS $= 2\left(\frac{7}{3}\right) + \frac{5}{3}$

LHS $= \frac{14}{3} + \frac{5}{3}$

Since the denominators are the same, we can add the numerators:

LHS $= \frac{14 + 5}{3}$

LHS $= \frac{19}{3}$

Right Hand Side (RHS):

RHS $= \frac{26}{3} - y$

Substitute $y = \frac{7}{3}$:

RHS $= \frac{26}{3} - \frac{7}{3}$

Since the denominators are the same, we can subtract the numerators:

RHS $= \frac{26 - 7}{3}$

RHS $= \frac{19}{3}$

Since LHS = RHS ($\frac{19}{3} = \frac{19}{3}$), the solution $y = \frac{7}{3}$ is correct.

Given:

The equation: $3m = 5m - \frac{8}{5}$

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To Solve:

Find the value of $m$.

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Solution:

We are given the equation:

$3m = 5m - \frac{8}{5}$

To eliminate the fraction, multiply both sides of the equation by $5$:

$5 \times 3m = 5 \times \left(5m - \frac{8}{5}\right)$

Distribute the $5$ on the right side:

$15m = 5 \times 5m - 5 \times \frac{8}{5}$

Simplify the equation:

$15m = 25m - 8$

Now, collect the terms containing $m$ on one side and the constant term on the other side.

Subtract $15m$ from both sides of the equation:

$15m - 15m = 25m - 8 - 15m$

Simplify both sides:

$0 = 10m - 8$

Now, add $8$ to both sides of the equation:

$0 + 8 = 10m - 8 + 8$

Simplify both sides:

$8 = 10m$

Now, divide both sides by $10$ to isolate $m$:

$\frac{8}{10} = \frac{10m}{10}$

Simplify the fraction on the left side:

$\frac{\cancel{8}^4}{\cancel{10}_5} = m$

So, the solution to the equation is $m = \frac{4}{5}$.

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Check:

To check the result, substitute $m = \frac{4}{5}$ back into the original equation $3m = 5m - \frac{8}{5}$.

Left Hand Side (LHS):

LHS $= 3m$

Substitute $m = \frac{4}{5}$:

LHS $= 3 \times \frac{4}{5}$

LHS $= \frac{12}{5}$

Right Hand Side (RHS):

RHS $= 5m - \frac{8}{5}$

Substitute $m = \frac{4}{5}$:

RHS $= 5\left(\frac{4}{5}\right) - \frac{8}{5}$

RHS $= \frac{\cancel{5}^1}{\cancel{5}_1} \times 4 - \frac{8}{5}$

RHS $= 4 - \frac{8}{5}$

To subtract the fraction from the integer, find a common denominator:

RHS $= \frac{4 \times 5}{5} - \frac{8}{5}$

RHS $= \frac{20}{5} - \frac{8}{5}$

RHS $= \frac{20 - 8}{5}$

RHS $= \frac{12}{5}$

Since LHS = RHS ($\frac{12}{5} = \frac{12}{5}$), the solution $m = \frac{4}{5}$ is correct.

Given:

The equation: $\frac{6x + 1}{3} + 1 = \frac{x - 3}{6}$

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To Solve:

Find the value of $x$ that satisfies the equation.

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Solution:

The given equation is:

$\frac{6x + 1}{3} + 1 = \frac{x - 3}{6}$

To remove the fractions, we multiply every term on both sides of the equation by the Least Common Multiple (LCM) of the denominators $3$ and $6$.

Let's find the LCM of 3 and 6:

$$\begin{array}{c|cc} 2 & 3 \;, & 6 \\ \hline 3 & 3 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$$

The LCM is $2 \times 3 = 6$.

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Multiply both sides of the equation by 6:

$6 \times \left(\frac{6x + 1}{3}\right) + 6 \times 1 = 6 \times \left(\frac{x - 3}{6}\right)$

Simplify the terms by cancelling common factors:

$\cancel{6}^{2} \times \frac{6x + 1}{\cancel{3}_1} + 6 = \cancel{6}^{1} \times \frac{x - 3}{\cancel{6}_1}$

This simplifies to:

$2(6x + 1) + 6 = 1(x - 3)$

Now, distribute the numbers outside the parentheses:

$12x + 2 + 6 = x - 3$

Combine the constant terms on the left side:

$12x + 8 = x - 3$

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Now, we collect the terms containing $x$ on one side and the constant terms on the other side.

Subtract $x$ from both sides of the equation:

$12x + 8 - x = x - 3 - x$

Simplify both sides:

$11x + 8 = -3$

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Subtract $8$ from both sides of the equation:

$11x + 8 - 8 = -3 - 8$

Simplify both sides:

$11x = -11$

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Divide both sides by $11$ to solve for $x$:

$\frac{11x}{11} = \frac{-11}{11}$

Simplify:

$x = -1$

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Check:

Substitute $x = -1$ back into the original equation $\frac{6x + 1}{3} + 1 = \frac{x - 3}{6}$.

Left Hand Side (LHS):

LHS $= \frac{6(-1) + 1}{3} + 1 = \frac{-6 + 1}{3} + 1 = \frac{-5}{3} + 1$

To add the fraction and the integer:

LHS $= -\frac{5}{3} + \frac{3}{3} = \frac{-5 + 3}{3} = \frac{-2}{3}$

Right Hand Side (RHS):

RHS $= \frac{-1 - 3}{6} = \frac{-4}{6}$

Simplify the fraction:

RHS $= \frac{\cancel{-4}^{-2}}{\cancel{6}_3} = -\frac{2}{3}$

Since LHS = RHS ($-\frac{2}{3} = -\frac{2}{3}$), the solution $x = -1$ is correct.

Thus, the solution is $\mathbf{x = -1}$.

Given:

The equation: $5x - 2 (2x - 7) = 2 (3x - 1) + \frac{7}{2}$

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To Solve:

Find the value of $x$ that satisfies the equation.

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Solution:

We are given the equation:

$5x - 2 (2x - 7) = 2 (3x - 1) + \frac{7}{2}$

First, simplify both sides of the equation by distributing the numbers outside the parentheses:

$5x - 4x + 14 = 6x - 2 + \frac{7}{2}$

Combine like terms on the left side and combine constant terms on the right side:

$x + 14 = 6x + \left(-2 + \frac{7}{2}\right)$

$x + 14 = 6x + \left(\frac{-4}{2} + \frac{7}{2}\right)$

$x + 14 = 6x + \frac{3}{2}$

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To eliminate the fraction, multiply both sides of the equation by $2$:

$2 \times (x + 14) = 2 \times \left(6x + \frac{3}{2}\right)$

Distribute the $2$ on both sides and simplify:

$2x + 28 = 2 \times 6x + \cancel{2} \times \frac{3}{\cancel{2}}$

$2x + 28 = 12x + 3$

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Now, collect the terms containing $x$ on one side and the constant terms on the other side.

Subtract $2x$ from both sides of the equation:

$2x + 28 - 2x = 12x + 3 - 2x$

Simplify both sides:

$28 = 10x + 3$

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Subtract $3$ from both sides of the equation:

$28 - 3 = 10x + 3 - 3$

Simplify both sides:

$25 = 10x$

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Divide both sides by $10$ to isolate $x$:

$\frac{25}{10} = \frac{10x}{10}$

Simplify the fraction:

$\frac{\cancel{25}^5}{\cancel{10}_2} = x$

So, $x = \frac{5}{2}$.

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Check:

Substitute $x = \frac{5}{2}$ back into the original equation $5x - 2 (2x - 7) = 2 (3x - 1) + \frac{7}{2}$.

Left Hand Side (LHS):

LHS $= 5\left(\frac{5}{2}\right) - 2 \left(2\left(\frac{5}{2}\right) - 7\right)$

LHS $= \frac{25}{2} - 2 \left(\cancel{2} \times \frac{5}{\cancel{2}} - 7\right)$

LHS $= \frac{25}{2} - 2 (5 - 7) = \frac{25}{2} - 2 (-2) = \frac{25}{2} + 4$

To add the fraction and the integer:

LHS $= \frac{25}{2} + \frac{8}{2} = \frac{25 + 8}{2} = \frac{33}{2}$

Right Hand Side (RHS):

RHS $= 2 \left(3\left(\frac{5}{2}\right) - 1\right) + \frac{7}{2}$

RHS $= 2 \left(\frac{15}{2} - 1\right) + \frac{7}{2}$

Inside the parentheses, find a common denominator:

RHS $= 2 \left(\frac{15}{2} - \frac{2}{2}\right) + \frac{7}{2} = 2 \left(\frac{13}{2}\right) + \frac{7}{2}$

Simplify the first term:

RHS $= \cancel{2} \times \frac{13}{\cancel{2}} + \frac{7}{2} = 13 + \frac{7}{2}$

To add the integer and the fraction:

RHS $= \frac{26}{2} + \frac{7}{2} = \frac{26 + 7}{2} = \frac{33}{2}$

Since LHS = RHS ($\frac{33}{2} = \frac{33}{2}$), the solution $x = \frac{5}{2}$ is correct.

Thus, the solution is $\mathbf{x = \frac{5}{2}}$.

Given:

The equation: $\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$

---

To Solve:

Find the value of $x$.

---

Solution:

We are given the equation:

$\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators $2, 5, 3,$ and $4$.

Let's find the LCM of 2, 5, 3, and 4:

$$\begin{array}{c|cccc} 2 & 2 \;, & 5 \;, & 3 \;, & 4 \\ \hline 2 & 1 \; , & 5 \; , & 3 \; , & 2 \\ \hline 3 & 1 \; , & 5 \; , & 3 \; , & 1 \\ \hline 5 & 1 \; , & 1 \; , & 1 \; , & 1 \\ \hline & 1 \; , & 1 \; , & 1 \; , & 1 \end{array}$$

LCM$(2, 5, 3, 4) = 2 \times 2 \times 3 \times 5 = 60$

---

Multiply both sides of the equation by the LCM, $60$:

$60 \times \left(\frac{x}{2} - \frac{1}{5}\right) = 60 \times \left(\frac{x}{3} + \frac{1}{4}\right)$

Distribute $60$ on both sides and simplify:

$60 \times \frac{x}{2} - 60 \times \frac{1}{5} = 60 \times \frac{x}{3} + 60 \times \frac{1}{4}$

$\cancel{60}^{30} \times \frac{x}{\cancel{2}_1} - \cancel{60}^{12} \times \frac{1}{\cancel{5}_1} = \cancel{60}^{20} \times \frac{x}{\cancel{3}_1} + \cancel{60}^{15} \times \frac{1}{\cancel{4}_1}$

$30x - 12 = 20x + 15$

---

Collect the terms containing $x$ on one side and the constant terms on the other side.

Subtract $20x$ from both sides:

$30x - 12 - 20x = 20x + 15 - 20x$

Simplify both sides:

$10x - 12 = 15$

Add $12$ to both sides:

$10x - 12 + 12 = 15 + 12$

Simplify both sides:

$10x = 27$

---

Divide both sides by $10$ to isolate $x$:

$\frac{10x}{10} = \frac{27}{10}$

Simplify both sides:

$x = \frac{27}{10}$

---

Check:

Substitute $x = \frac{27}{10}$ into $\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$.

Left Hand Side (LHS):

LHS $= \frac{27/10}{2} - \frac{1}{5} = \frac{27}{20} - \frac{1}{5}$

Find a common denominator (20):

LHS $= \frac{27}{20} - \frac{1 \times 4}{5 \times 4} = \frac{27}{20} - \frac{4}{20} = \frac{27 - 4}{20} = \frac{23}{20}$

Right Hand Side (RHS):

RHS $= \frac{27/10}{3} + \frac{1}{4} = \frac{27}{30} + \frac{1}{4}$

Simplify $\frac{27}{30}$ and find a common denominator (20):

RHS $= \frac{\cancel{27}^9}{\cancel{30}_{10}} + \frac{1}{4} = \frac{9}{10} + \frac{1}{4}$

RHS $= \frac{9 \times 2}{10 \times 2} + \frac{1 \times 5}{4 \times 5} = \frac{18}{20} + \frac{5}{20} = \frac{18 + 5}{20} = \frac{23}{20}$

Since LHS = RHS, the solution $x = \frac{27}{10}$ is correct.

Thus, the solution is $\mathbf{x = \frac{27}{10}}$.

Given:

The equation: $\frac{n}{2} - \frac{3n}{4} + \frac{5n}{6} = 21$

---

To Solve:

Find the value of $n$.

---

Solution:

We are given the equation:

$\frac{n}{2} - \frac{3n}{4} + \frac{5n}{6} = 21$

To clear the fractions, we find the Least Common Multiple (LCM) of the denominators $2$, $4$, and $6$.

Let's find the LCM of 2, 4, and 6:

$$\begin{array}{c|ccc} 2 & 2 \;, & 4 \;, & 6 \\ \hline 2 & 1 \; , & 2 \; , & 3 \\ \hline 3 & 1 \; , & 1 \; , & 3 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$$

The LCM of $2, 4, 6$ is $2 \times 2 \times 3 = 12$.

---

Multiply every term on both sides by the LCM, $12$:

$12 \times \frac{n}{2} - 12 \times \frac{3n}{4} + 12 \times \frac{5n}{6} = 12 \times 21$

Simplify each term by performing cancellation:

$\cancel{12}^{6} \times \frac{n}{\cancel{2}_1} - \cancel{12}^{3} \times \frac{3n}{\cancel{4}_1} + \cancel{12}^{2} \times \frac{5n}{\cancel{6}_1} = 12 \times 21$

$6n - 3(3n) + 2(5n) = 252$

$6n - 9n + 10n = 252$

---

Combine the terms involving $n$ on the left side:

$(6 - 9 + 10)n = 252$

$7n = 252$

---

Now, divide both sides by $7$ to isolate $n$:

$\frac{7n}{7} = \frac{252}{7}$

Simplify both sides:

$n = 36$

---

Check:

Substitute $n = 36$ back into the original equation $\frac{n}{2} - \frac{3n}{4} + \frac{5n}{6} = 21$.

Left Hand Side (LHS):

LHS $= \frac{36}{2} - \frac{3(36)}{4} + \frac{5(36)}{6} = 18 - \frac{108}{4} + \frac{180}{6}$

Perform the divisions:

LHS $= 18 - 27 + 30$

Combine the terms:

LHS $= (18 + 30) - 27 = 48 - 27 = 21$

Right Hand Side (RHS):

RHS $= 21$

Since LHS = RHS, the solution $n = 36$ is correct.

Thus, the solution is $\mathbf{n = 36}$.

Given:

The equation: $x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{2}$

---

To Solve:

Find the value of $x$.

---

Solution:

We are given the equation:

$x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{2}$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators $3$, $6$, and $2$.

Let's find the LCM of $3, 6,$ and $2$:

$$\begin{array}{c|ccc} 2 & 3 \;, & 6 \;, & 2 \\ \hline 3 & 3 \; , & 3 \; , & 1 \\ \hline & 1 \; , & 1 \; , & 1 \end{array}$$

The LCM of $3, 6,$ and $2$ is $2 \times 3 = 6$.

---

Multiply every term on both sides of the equation by the LCM, $6$:

$6 \times x + 6 \times 7 - 6 \times \frac{8x}{3} = 6 \times \frac{17}{6} - 6 \times \frac{5x}{2}$

Simplify each term by performing cancellation:

$6x + 42 - \cancel{6}^{2} \times \frac{8x}{\cancel{3}_1} = \cancel{6}^{1} \times \frac{17}{\cancel{6}_1} - \cancel{6}^{3} \times \frac{5x}{\cancel{2}_1}$

Perform the multiplications:

$6x + 42 - 16x = 17 - 15x$

---

Combine the terms involving $x$ on the left side:

$(6x - 16x) + 42 = 17 - 15x$

$-10x + 42 = 17 - 15x$

---

Collect the terms containing $x$ on one side and the constant terms on the other side.

Add $15x$ to both sides:

$-10x + 42 + 15x = 17 - 15x + 15x$

Simplify both sides:

$5x + 42 = 17$

Subtract $42$ from both sides:

$5x + 42 - 42 = 17 - 42$

Simplify both sides:

$5x = -25$

---

Divide both sides by $5$ to isolate $x$:

$\frac{5x}{5} = \frac{-25}{5}$

Simplify both sides:

$x = -5$

---

Check:

Substitute $x = -5$ back into $x + 7 - \frac{8x}{3} = \frac{17}{6} - \frac{5x}{2}$.

Left Hand Side (LHS):

LHS $= -5 + 7 - \frac{8(-5)}{3} = 2 - \frac{-40}{3} = 2 + \frac{40}{3}$

Find a common denominator (3):

LHS $= \frac{2 \times 3}{3} + \frac{40}{3} = \frac{6}{3} + \frac{40}{3} = \frac{6 + 40}{3} = \frac{46}{3}$

Right Hand Side (RHS):

RHS $= \frac{17}{6} - \frac{5(-5)}{2} = \frac{17}{6} - \frac{-25}{2} = \frac{17}{6} + \frac{25}{2}$

Find a common denominator (6):

RHS $= \frac{17}{6} + \frac{25 \times 3}{2 \times 3} = \frac{17}{6} + \frac{75}{6} = \frac{17 + 75}{6} = \frac{92}{6}$

Simplify the fraction:

RHS $= \frac{\cancel{92}^{46}}{\cancel{6}_3} = \frac{46}{3}$

Since LHS = RHS, the solution $x = -5$ is correct.

Thus, the solution is $\mathbf{x = -5}$.

Given:

The equation: $\frac{x - 5}{3} = \frac{x - 3}{5}$

---

To Solve:

Find the value of $x$.

---

Solution:

We are given the equation:

$\frac{x - 5}{3} = \frac{x - 3}{5}$

We can solve this by cross-multiplication.

$5 \times (x - 5) = 3 \times (x - 3)$

Now, distribute the numbers on both sides:

$5x - 25 = 3x - 9$

---

Collect the terms containing $x$ on one side and the constant terms on the other side.

Subtract $3x$ from both sides:

$5x - 25 - 3x = 3x - 9 - 3x$

Simplify both sides:

$2x - 25 = -9$

Add $25$ to both sides:

$2x - 25 + 25 = -9 + 25$

Simplify both sides:

$2x = 16$

---

Divide both sides by $2$ to isolate $x$:

$\frac{2x}{2} = \frac{16}{2}$

Simplify both sides:

$x = 8$

---

Check:

Substitute $x = 8$ back into $\frac{x - 5}{3} = \frac{x - 3}{5}$.

Left Hand Side (LHS):

LHS $= \frac{8 - 5}{3} = \frac{3}{3} = 1$

Right Hand Side (RHS):

RHS $= \frac{8 - 3}{5} = \frac{5}{5} = 1$

Since LHS = RHS, the solution $x = 8$ is correct.

Thus, the solution is $\mathbf{x = 8}$.

Given:

The equation: $\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t$

---

To Solve:

Find the value of $t$.

---

Solution:

We are given the equation:

$\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators $4, 3,$ and $3$. Let's find the LCM of $4$ and $3$:

$$\begin{array}{c|cc} 2 & 4 \;, & 3 \\ \hline 2 & 2 \; , & 3 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$$

The LCM of $4$ and $3$ is $2 \times 2 \times 3 = 12$.

---

Multiply every term on both sides of the equation by the LCM, $12$:

$12 \times \left(\frac{3t - 2}{4}\right) - 12 \times \left(\frac{2t + 3}{3}\right) = 12 \times \frac{2}{3} - 12 \times t$

Simplify each term by performing cancellation:

$\cancel{12}^{3} \times \frac{3t - 2}{\cancel{4}_1} - \cancel{12}^{4} \times \frac{2t + 3}{\cancel{3}_1} = \cancel{12}^{4} \times \frac{2}{\cancel{3}_1} - 12t$

This simplifies to:

$3(3t - 2) - 4(2t + 3) = 8 - 12t$

---

Now, distribute the numbers outside the parentheses. Be careful with the signs:

$(9t - 6) - (8t + 12) = 8 - 12t$

Remove the parentheses:

$9t - 6 - 8t - 12 = 8 - 12t$

Combine the terms involving $t$ and the constant terms on the left side:

$(9t - 8t) + (-6 - 12) = 8 - 12t$

$t - 18 = 8 - 12t$

---

Collect the terms containing $t$ on one side and the constant terms on the other side.

Add $12t$ to both sides:

$t - 18 + 12t = 8 - 12t + 12t$

Combine like terms:

$13t - 18 = 8$

Add $18$ to both sides:

$13t - 18 + 18 = 8 + 18$

Simplify both sides:

$13t = 26$

---

Divide both sides by $13$ to isolate $t$:

$\frac{13t}{13} = \frac{26}{13}$

Simplify both sides:

$t = 2$

---

Check:

Substitute $t = 2$ back into $\frac{3t - 2}{4} - \frac{2t + 3}{3} = \frac{2}{3} - t$.

Left Hand Side (LHS):

LHS $= \frac{3(2) - 2}{4} - \frac{2(2) + 3}{3} = \frac{6 - 2}{4} - \frac{4 + 3}{3} = \frac{4}{4} - \frac{7}{3} = 1 - \frac{7}{3}$

Find a common denominator (3):

LHS $= \frac{3}{3} - \frac{7}{3} = \frac{3 - 7}{3} = \frac{-4}{3}$

Right Hand Side (RHS):

RHS $= \frac{2}{3} - 2$

Find a common denominator (3):

RHS $= \frac{2}{3} - \frac{2 \times 3}{3} = \frac{2}{3} - \frac{6}{3} = \frac{2 - 6}{3} = \frac{-4}{3}$

Since LHS = RHS, the solution $t = 2$ is correct.

Thus, the solution is $\mathbf{t = 2}$.

Given:

The equation: $m - \frac{m - 1}{2} = 1 - \frac{m - 2}{3}$

---

To Solve:

Find the value of $m$ that satisfies the equation.

---

Solution:

We are given the equation:

$m - \frac{m - 1}{2} = 1 - \frac{m - 2}{3}$

To eliminate the fractions, we find the Least Common Multiple (LCM) of the denominators $2$ and $3$.

Let's find the LCM of 2 and 3:

$$\begin{array}{c|cc} 2 & 2 \;, & 3 \\ \hline 3 & 1 \; , & 3 \\ \hline & 1 \; , & 1 \end{array}$$

The LCM of $2$ and $3$ is $2 \times 3 = 6$.

---

Multiply every term on both sides of the equation by the LCM, $6$:

$6 \times m - 6 \times \left(\frac{m - 1}{2}\right) = 6 \times 1 - 6 \times \left(\frac{m - 2}{3}\right)$

Simplify each term by performing cancellation:

$6m - \cancel{6}^{3} \times \frac{m - 1}{\cancel{2}_1} = 6 - \cancel{6}^{2} \times \frac{m - 2}{\cancel{3}_1}$

This simplifies to:

$6m - 3(m - 1) = 6 - 2(m - 2)$

---

Now, distribute the numbers outside the parentheses. Be careful with the signs:

$(6m - 3m) + 3 = 6 - (2m - 4)$

$3m + 3 = 6 - 2m + 4$

Combine constants on the RHS:

$3m + 3 = 10 - 2m$

---

Collect the terms containing $m$ on one side and the constant terms on the other side.

Add $2m$ to both sides:

$3m + 3 + 2m = 10 - 2m + 2m$

Combine like terms:

$5m + 3 = 10$

Subtract $3$ from both sides:

$5m + 3 - 3 = 10 - 3$

Simplify both sides:

$5m = 7$

---

Divide both sides by $5$ to isolate $m$:

$\frac{5m}{5} = \frac{7}{5}$

Simplify both sides:

$m = \frac{7}{5}$

---

Check:

Substitute $m = \frac{7}{5}$ back into $m - \frac{m - 1}{2} = 1 - \frac{m - 2}{3}$.

Left Hand Side (LHS):

LHS $= \frac{7}{5} - \frac{\frac{7}{5} - 1}{2} = \frac{7}{5} - \frac{\frac{7-5}{5}}{2} = \frac{7}{5} - \frac{\frac{2}{5}}{2} = \frac{7}{5} - \left(\frac{2}{5} \times \frac{1}{2}\right) = \frac{7}{5} - \frac{\cancel{2}^1}{\cancel{10}_5} = \frac{7}{5} - \frac{1}{5} = \frac{6}{5}$

Right Hand Side (RHS):

RHS $= 1 - \frac{\frac{7}{5} - 2}{3} = 1 - \frac{\frac{7-10}{5}}{3} = 1 - \frac{-\frac{3}{5}}{3} = 1 - \left(\frac{-3}{5} \times \frac{1}{3}\right) = 1 - \frac{-\cancel{3}^1}{\cancel{15}_5} = 1 - \left(-\frac{1}{5}\right) = 1 + \frac{1}{5} = \frac{5}{5} + \frac{1}{5} = \frac{6}{5}$

Since LHS = RHS, the solution $m = \frac{7}{5}$ is correct.

Thus, the solution is $\mathbf{m = \frac{7}{5}}$.

Given:

The equation: $3(t - 3) = 5(2t + 1)$

---

To Solve:

Find the value of $t$ that satisfies the equation.

---

Solution:

We are given the equation:

$3(t - 3) = 5(2t + 1)$

First, simplify both sides of the equation by distributing the numbers outside the parentheses:

$3t - 9 = 10t + 5$

---

Collect the terms containing $t$ on one side and the constant terms on the other side.

Subtract $3t$ from both sides:

$3t - 9 - 3t = 10t + 5 - 3t$

Simplify both sides:

$-9 = 7t + 5$

Subtract $5$ from both sides:

$-9 - 5 = 7t + 5 - 5$

Simplify both sides:

$-14 = 7t$

---

Divide both sides by $7$ to isolate $t$:

$\frac{-14}{7} = \frac{7t}{7}$

Simplify both sides:

$-2 = t$

---

Check:

Substitute $t = -2$ back into $3(t - 3) = 5(2t + 1)$.

Left Hand Side (LHS):

LHS $= 3(-2 - 3) = 3(-5) = -15$

Right Hand Side (RHS):

RHS $= 5(2(-2) + 1) = 5(-4 + 1) = 5(-3) = -15$

Since LHS = RHS, the solution $t = -2$ is correct.

Thus, the solution is $\mathbf{t = -2}$.

Given:

The equation: $15(y - 4) - 2(y - 9) + 5(y + 6) = 0$

---

To Solve:

Find the value of $y$ that satisfies the equation.

---

Solution:

We are given the equation:

$15(y - 4) - 2(y - 9) + 5(y + 6) = 0$

First, simplify the left side by distributing the numbers outside the parentheses:

$15y - 60 - (2y - 18) + (5y + 30) = 0$

Remove the parentheses:

$15y - 60 - 2y + 18 + 5y + 30 = 0$

---

Combine the terms involving $y$ and the constant terms on the left side:

$(15y - 2y + 5y) + (-60 + 18 + 30) = 0$

$18y + (-12) = 0$

$18y - 12 = 0$

---

Isolate the term containing $y$. Add $12$ to both sides:

$18y - 12 + 12 = 0 + 12$

Simplify both sides:

$18y = 12$

---

Divide both sides by $18$ to isolate $y$:

$\frac{18y}{18} = \frac{12}{18}$

Simplify the fraction:

$y = \frac{\cancel{12}^2}{\cancel{18}_3}$

---

Check:

Substitute $y = \frac{2}{3}$ back into $15(y - 4) - 2(y - 9) + 5(y + 6) = 0$.

Left Hand Side (LHS):

LHS $= 15\left(\frac{2}{3} - 4\right) - 2\left(\frac{2}{3} - 9\right) + 5\left(\frac{2}{3} + 6\right)$

Simplify inside parentheses:

LHS $= 15\left(\frac{2-12}{3}\right) - 2\left(\frac{2-27}{3}\right) + 5\left(\frac{2+18}{3}\right)$

LHS $= 15\left(\frac{-10}{3}\right) - 2\left(\frac{-25}{3}\right) + 5\left(\frac{20}{3}\right)$

Perform multiplications:

LHS $= \frac{15 \times -10}{3} - \frac{2 \times -25}{3} + \frac{5 \times 20}{3} = \frac{-150}{3} + \frac{50}{3} + \frac{100}{3}$

Combine numerators:

LHS $= \frac{-150 + 50 + 100}{3} = \frac{0}{3} = 0$

Right Hand Side (RHS):

RHS $= 0$

Since LHS = RHS, the solution $y = \frac{2}{3}$ is correct.

Thus, the solution is $\mathbf{y = \frac{2}{3}}$.

Given:

The equation: $3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17$

---

To Solve:

Find the value of $z$ that satisfies the equation.

---

Solution:

We are given the equation:

$3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17$

First, simplify both sides by distributing the numbers outside the parentheses:

$(15z - 21) - (18z - 22) = (32z - 52) - 17$

Remove the parentheses:

$15z - 21 - 18z + 22 = 32z - 52 - 17$

---

Combine the terms involving $z$ and the constant terms on each side:

$(15z - 18z) + (-21 + 22) = 32z + (-52 - 17)$

$-3z + 1 = 32z - 69$

---

Collect the terms containing $z$ on one side and the constant terms on the other side.

Add $3z$ to both sides:

$-3z + 1 + 3z = 32z - 69 + 3z$

Simplify both sides:

$1 = 35z - 69$

Add $69$ to both sides:

$1 + 69 = 35z - 69 + 69$

Simplify both sides:

$70 = 35z$

---

Divide both sides by $35$ to isolate $z$:

$\frac{70}{35} = \frac{35z}{35}$

Simplify both sides:

$2 = z$

---

Check:

Substitute $z = 2$ back into $3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17$.

Left Hand Side (LHS):

LHS $= 3(5(2) - 7) - 2(9(2) - 11) = 3(10 - 7) - 2(18 - 11) = 3(3) - 2(7) = 9 - 14 = -5$

Right Hand Side (RHS):

RHS $= 4(8(2) - 13) - 17 = 4(16 - 13) - 17 = 4(3) - 17 = 12 - 17 = -5$

Since LHS = RHS, the solution $z = 2$ is correct.

Thus, the solution is $\mathbf{z = 2}$.

Given:

The equation: $0.25(4f - 3) = 0.05(10f - 9)$

---

To Solve:

Find the value of $f$.

---

Solution:

We are given the equation:

$0.25(4f - 3) = 0.05(10f - 9)$

First, simplify both sides by distributing the numbers outside the parentheses:

$0.25 \times 4f - 0.25 \times 3 = 0.05 \times 10f - 0.05 \times 9$

Perform the multiplications:

$f - 0.75 = 0.5f - 0.45$

---

Collect the terms containing $f$ on one side and the constant terms on the other side.

Subtract $0.5f$ from both sides:

$f - 0.75 - 0.5f = 0.5f - 0.45 - 0.5f$

Combine like terms:

$0.5f - 0.75 = -0.45$

Add $0.75$ to both sides:

$0.5f - 0.75 + 0.75 = -0.45 + 0.75$

Simplify both sides:

$0.5f = 0.30$

---

Divide both sides by $0.5$ to isolate $f$:

$\frac{0.5f}{0.5} = \frac{0.30}{0.5}$

Simplify the division:

$f = \frac{0.3}{0.5} = \frac{3}{5} = 0.6$

---

Check:

Substitute $f = 0.6$ back into $0.25(4f - 3) = 0.05(10f - 9)$.

Left Hand Side (LHS):

LHS $= 0.25(4(0.6) - 3) = 0.25(2.4 - 3) = 0.25(-0.6) = -0.15$

Right Hand Side (RHS):

RHS $= 0.05(10(0.6) - 9) = 0.05(6 - 9) = 0.05(-3) = -0.15$

Since LHS = RHS, the solution $f = 0.6$ is correct.

Thus, the solution is $\mathbf{f = 0.6}$.