# Solutions to Equations - Grade 9

The solutions to the questions in solve equations are presented along with all the steps needed and detailed explanations.

## Solutions to Questions in solve equations

1. )
Given
$\quad \quad 2x + 2 = 6$
Subtract $2$ from both sides of the equation
$\quad \quad 2x + 2 - 2 = 6 - 2$
Group like terms and simplify
$\quad \quad 2x = 4$
Divide both sides of the equation by $2$ (or multiply both sides by \dfrac{1}{2} \)
$\quad \quad \dfrac{2x}{2} = \dfrac{4}{2}$
Simplify
$\quad \quad x = 2$
Check the solution obtained
Left side of equation for $x = 2$ : $\quad 2x + 2 = 2(2)+2 = 6$
Left side and right are both equal to $6$ for $x = 2$, therefore $x = 2$ is a solution to the given equation.

2. )
Given the equation
$\quad \quad 5y - 2 = 7y - 8$
Let us arrange the equation so that the variable is on the right side and the constant is on the left side. Subtract $5y$ from both sides.
$\quad \quad 5y - 2 - 5 y = 7y - 8 - 5 y$
Group like terms and simplify
$\quad \quad - 2 = 2y - 8$
Add $8$ to both sides.
$\quad \quad - 2 + 8 = 2y - 8 + 8$
Group like terms and simplify
$\quad \quad 6 = 2y$
Divide both sides of the equation by $2$ (or multiply both sides by $\dfrac{1}{2}$ ).
$\quad \quad \dfrac{6}{2} = \dfrac{2y}{2}$
Simplify
$\quad \quad 3 = y$
Check the solution obtained
Left side of equation for $y = 3$ : $\quad 5y - 2 = 5(3) - 2 = 13$
Right side of equation for $y = 3$ : $\quad 7y - 8 = 7(3) - 8 = 13$
Hence $y = 3$ is a solution to the given equation.

3. )
Given
$\quad \quad -2x + 4 + 5x = 7 + 4x - 3$
Group like terms and simplify
$\quad \quad 3x + 4 = 4 + 4x$

Subtract $4$ and $3x$ from both sides. (Two opeartions in one step).
$\quad \quad 3x + 4 - 4 - 3x = 4 + 4x - 4 - 3x$
Group like terms and simplify
$\quad \quad 0 = x$
Check the solution obtained
Left side of equation for $x = 0$ : $\quad -2x + 4 + 5x = -2(0) + 4 + 5(0) = 4$
Right side of equation for $x = 0$ : $\quad 7 + 4x - 3 = 7 + 4(0) - 3 = 4$
Hence $x = 0$ is a solution to the given equation.

4. )
Given
$\quad \quad 0.2 d + 4 = - 0.1 d - 2$
Add $0.1d$ to both sides and subtract $4$ from both sides. (Two opeartions in one step).
$\quad \quad 0.2 d + 4 + 0.1 d - 4 = - 0.1 d - 2 + 0.1 d - 4$
Group like terms and simplify
$\quad \quad 0.3 d = - 6$
Divide both sides by $0.3$ and simplify
$\quad \quad \dfrac{0.3 d}{0.3} = - \dfrac{6}{0.3}$
$\quad \quad d = - 20$
Check the solution obtained
Left side of equation for $d = -20$ : $\quad 0.2 d + 4 = 0.2(-20) + 4 = 0$
Right side of equation for $d = - 20$ : $\quad - 0.1 d - 2 = - 0.1 (-20) - 2 = 0$
Hence $d = -20$ is a solution to the given equation.

5. )
Given the equation
$\quad \quad -2(2x- 6) = -(x - 4)$
Use distributive law to expand both sides.
$\quad \quad -4 x + 12 = - x + 4$
Add $x$ to both sides, group like terms and simplify
$\quad \quad -4 x + 12 + x = - x + 4 + x$
$\quad \quad -3 x + 12 = 4$
Subtract $12$ from both sides, group like terms and simplify
$\quad \quad -3 x + 12 - 12 = 4 - 12$
$\quad \quad -3 x = - 8$
Divide both sides by $- 3$ and simplify
$\quad \quad - \dfrac{ 3 x}{-3} = - \dfrac{8}{-3}$
$\quad \quad x = \dfrac{8}{3}$
Check the solution obtained
Left side of equation for $x = \dfrac{8}{3}$ : $\quad -2(2x- 6) = -2(2 (\dfrac{8}{3})- 6) = \dfrac{4}{3}$
Right side of equation for $x = \dfrac{8}{3}$ : $\quad -(\dfrac{8}{3} - 4) = \dfrac{4}{3}$
Hence $x = \dfrac{8}{3}$ is a solution to the given equation.

6. )
Given
$\quad \quad -(x+2)+4 = 2(x+3) + x$
Use distributive law to expand both sides, group like terms and simplify.
$\quad \quad -x - 2 +4 = 2x + 6 + x$
$\quad \quad -x + 2 = 3x + 6$
Add $x$ to both sides and subtract $6$ from both sides, group like terms and simplify.
$\quad \quad -x + 2 + x - 6 = 3x + 6 + x - 6$
$\quad \quad - 4 = 4 x$
Divide both sides by $4$ to obtain the solution
$\quad \quad x = - 1$
Check the solution obtained
Left side of equation for $x = - 1$ : $\quad -(x+2)+4 = -(-1+2)+4 = 3$
Right side of equation for $x = - 1$ : $\quad = 2(x+3) + x = 2(-1+3) - 1 = 3$
Hence $x = - 1$ is a solution to the given equation.

7. )
Given
$\quad \quad \dfrac{x}{5} = - 6$
In order to get rid of the fraction, multiply both sides of the equation by the denominator $5$
$\quad \quad 5 \left( \dfrac{x}{5} \right) = 5(- 6)$
Rearrange as
$\quad \quad \dfrac{5}{5} x = 5(- 6)$
Simplify to find the solution.
$\quad \quad x = - 30$
Check the solution obtained
Left side of equation for $x = - 30$ : $\quad \dfrac{x}{5} = \dfrac{-30}{5} = - 6$
Right side of equation is equal to $- 6$.
Hence $x = - 30$ is a solution to the given equation.

8. )
Given the equation
$\quad \quad - \dfrac{x}{3} = \dfrac{1}{2}$
Multiply both sides by $- 3$
$\quad \quad - 3 \left(- \dfrac{x}{3} \right) = - 3 \left(\dfrac{1}{2} \right)$
Simplify
$\quad \quad \dfrac{3}{3} x = \dfrac{-3}{2}$
$\quad \quad x = - \dfrac{3}{2}$
Check the solution obtained
Left side of equation for $x = - \dfrac{3}{2}$ : $\quad - \dfrac{x}{3} = - \dfrac{1}{3} x = - \dfrac{1}{3} \left( - \dfrac{3}{2} \right) = \dfrac{1}{2}$
Both sides of the equation are equal for $x = - \dfrac{3}{2}$
Hence $x = - \dfrac{3}{2}$ is a solution to the given equation.

9. )
Given
$\quad \quad - \dfrac{x}{4} = \dfrac{1}{2} - x$
Multiply both sides by the lowest common multiple (LCM) of the denominators $4$ and $2$ which is equal to $4$.
$\quad \quad 4 \left(- \dfrac{x}{4} \right) = 4 \left(\dfrac{1}{2} - x\right)$
Use distributive law to expand the right side of the equation
$\quad \quad 4 \left(- \dfrac{x}{4} \right) = 4 \left(\dfrac{1}{2} \right) - 4 x$
Rearrange the fractions as
$\quad \quad - \dfrac{4}{4} x = \dfrac{4}{2} - 4 x$
Simplify
$\quad \quad - x = 2 - 4 x$
Add $4x$ to both sides of the equation and group like terms
$\quad \quad - x + 4 x = 2 - 4 x + 4 x$
$\quad \quad 3x = 2$
Divide boh sides by $3$ and simplify to find the solution.
$\quad \quad x = \dfrac{2}{3}$
Check the solution obtained
Left side of equation for $x = \dfrac{2}{3}$ : $\quad - \dfrac{x}{4} = - \dfrac{1}{4} (x) = - \dfrac{1}{4} \left( \dfrac{2}{3} \right) = - \dfrac{1}{6}$

Right side of equation for $x = \dfrac{2}{3}$ : $\quad \dfrac{1}{2} - x = \dfrac{1}{2} - \dfrac{2}{3} = - \dfrac{1}{6}$
Both sides of the equation are equal for $x = \dfrac{2}{3}$
Hence $x = \dfrac{2}{3}$ is a solution to the given equation.

10. )
Given the equation
$\quad \quad - \dfrac{x-3}{7} = \dfrac{1}{2} (- 2x + 6)$
Find the lowest common multiple (LCM) of the denominators $7$ and $2$ which is equal to $14$ and multiply both sides of the equation by $14$.
$\quad \quad 14 \left(- \dfrac{x-3}{7} \right) = 14 \left(\dfrac{1}{2} (- 2x + 6) \right)$
Rearrange fractions in order to make it easier to simplify
$\quad \quad - \dfrac{14}{7} (x - 3) = \dfrac{14}{2} (- 2x + 6)$
Reduce the fractions $14/7 = 2$ and $14/2 = 7$
$\quad \quad - 2 (x - 3) = 7(- 2x + 6)$
Use distibutive law to expand both sides
$\quad \quad -2x + 6 = - 14 x + 42$
Subtract $6$ from both sides
$\quad \quad -2x + 6 - 6 = - 14 x + 42 - 6$
Group like terms and simplify
$\quad \quad - 2 x = - 14 x + 36$
Add $14 x$ to both sides
$\quad \quad - 2 x + 14 x = - 14 x + 36 + 14 x$
Group like terms and simplify
$\quad \quad 12 x = 36$
Divide both sides by $12$
$\quad \quad \dfrac{12x}{12} x = \dfrac{36}{12}$
Simplify
$\quad \quad x = 3$
Check the solution obtained
Left side of equation for $x = 3$ : $\quad - \dfrac{x-3}{7} = - \dfrac{3-3}{7} = 0$
Right side of equation for $x = 3$ : $\quad \dfrac{1}{2} (- 2x + 6) = \dfrac{1}{2} (- 2(3) + 6) = 0$
Both sides of the equation are equal for $x = 3$
Therefore $x = 3$ is a solution to the given equation.

11. )
Given
$\quad \quad - \dfrac{1}{2} - x + 5 = \dfrac{1}{5} + 2(x-2)$
Find the lowest common multiple (LCM) of the denominators $2$ and $5$ which is equal to $10$ and multiply both sides of the equation by $10$.
$\quad \quad 10 \left(- \dfrac{1}{2} - x + 5 \right) = 10 \left( \dfrac{1}{5} + 2(x-2) \right)$
Use distributive law to exppand both sides of the equation
$\quad \quad - 10 \left( \dfrac{1}{2} \right) -10x + 50 = 10 \left( \dfrac{1}{5} \right) + 20(x - 2)$

Expand $20(x - 2) = 2x - 40$ and simplify the fractions $10 \left( \dfrac{1}{2} \right) = 10/2 = 5$ and $10 \left( \dfrac{1}{5} \right) = 10/5 = 2$

$\quad \quad -5 - 10x + 50 = 2 + 20 x - 40$
Group like terms and simplify
$\quad \quad - 10 x + 45 = 20 x - 38$
Add $10 x$ from both sides
$\quad \quad - 10 x + 45 + 10 x = 20 x - 38 + 10 x$
Group like terms and simplify
$\quad \quad 45 = 30x - 38$
Add $38$ to both sides
$\quad \quad 45 + 38 = 30 x$
Group like terms and simplify
$\quad \quad 83 = 30x$
Divide both sides by $30$ and simplify to solve for $x$.
$\quad \quad x = \dfrac{83}{30}$

Check the solution obtained
Left side of equation for $x = \dfrac{83}{30}$ : $\quad - \dfrac{1}{2} - x + 5 = - \dfrac{1}{2} - \dfrac{83}{30} + 5 = 26/15$

Right side of equation for $x = \dfrac{83}{30}$ : $\quad \dfrac{1}{5} + 2(x-2) = \dfrac{1}{5} + 2(\dfrac{83}{30}-2) = 26/15$

Both sides of the equation are equal for $x = \dfrac{83}{30}$
Therefore $x = \dfrac{83}{30}$ is a solution to the given equation.