Solving Systems of Equations by Substitution worksheet with ten algebra problems.

Here are the step-by-step solutions for each system of equations using the substitution method.

1)
Given:
$y = \frac{7}{4}x - 3$
$y = 4$

Substitute $4$ for $y$ in the first equation:
$4 = \frac{7}{4}x - 3$
Add $3$ to both sides:
$7 = \frac{7}{4}x$
Multiply both sides by $\frac{4}{7}$:
$x = 4$

Solution: $(4, 4)$

2)
Given:
$y = -\frac{1}{4}x + 6$
$y = 4$

Substitute $4$ for $y$ in the first equation:
$4 = -\frac{1}{4}x + 6$
Subtract $6$ from both sides:
$-2 = -\frac{1}{4}x$
Multiply both sides by $-4$:
$x = 8$

Solution: $(8, 4)$

3)
Given:
$6x - 5y = 22$
$y = -8$

Substitute $-8$ for $y$ in the first equation:
$6x - 5(-8) = 22$
$6x + 40 = 22$
Subtract $40$ from both sides:
$6x = -18$
Divide by $6$:
$x = -3$

Solution: $(-3, -8)$

4)
Given:
$4x - 7y = 5$
$9x - 7y = -15$

Since both equations have $-7y$, we can solve the first one for $7y$ or subtract them. Let's isolate $7y$ in the first equation:
$4x - 5 = 7y$

Substitute $(4x - 5)$ for $7y$ in the second equation ($9x - (7y) = -15$):
$9x - (4x - 5) = -15$
$9x - 4x + 5 = -15$
$5x + 5 = -15$
Subtract $5$ from both sides:
$5x = -20$
Divide by $5$:
$x = -4$

Now find $y$ using $7y = 4x - 5$:
$7y = 4(-4) - 5$
$7y = -16 - 5$
$7y = -21$
$y = -3$

Solution: $(-4, -3)$

5)
Given:
$-5x + 9y = -12$
$3x + 2y = 22$

Solve the second equation for $x$:
$3x = 22 - 2y$
$x = \frac{22 - 2y}{3}$

Substitute this into the first equation:
$-5(\frac{22 - 2y}{3}) + 9y = -12$
Multiply everything by $3$ to clear the fraction:
$-5(22 - 2y) + 27y = -36$
$-110 + 10y + 27y = -36$
$37y - 110 = -36$
Add $110$ to both sides:
$37y = 74$
Divide by $37$:
$y = 2$

Find $x$:
$x = \frac{22 - 2(2)}{3} = \frac{18}{3} = 6$

Solution: $(6, 2)$

6)
Given:
$y = -\frac{2}{5}x - 2$
$y = -4$

Substitute $-4$ for $y$:
$-4 = -\frac{2}{5}x - 2$
Add $2$ to both sides:
$-2 = -\frac{2}{5}x$
Multiply by $-\frac{5}{2}$:
$x = 5$

Solution: $(5, -4)$

7)
Given:
$y = -\frac{3}{4}x + 4$
$y = -\frac{1}{2}x - 1$

Set the expressions for $y$ equal to each other:
$-\frac{3}{4}x + 4 = -\frac{1}{2}x - 1$
Add $\frac{3}{4}x$ to both sides and add $1$ to both sides:
$5 = \frac{3}{4}x - \frac{1}{2}x$
Convert $\frac{1}{2}$ to $\frac{2}{4}$:
$5 = \frac{3}{4}x - \frac{2}{4}x$
$5 = \frac{1}{4}x$
Multiply by $4$:
$x = 20$

Find $y$:
$y = -\frac{1}{2}(20) - 1 = -10 - 1 = -11$

Solution: $(20, -11)$

8)
Given:
$y = -\frac{4}{9}x - 3$
$y = -\frac{7}{5}x - 3$

Set them equal:
$-\frac{4}{9}x - 3 = -\frac{7}{5}x - 3$
Add $3$ to both sides:
$-\frac{4}{9}x = -\frac{7}{5}x$
The only number that makes this true is $0$.
$x = 0$

Find $y$:
$y = -\frac{4}{9}(0) - 3 = -3$

Solution: $(0, -3)$

9)
Given:
$y = \frac{3}{5}x + 2$
$y = -\frac{1}{3}x + 2$

Set them equal:
$\frac{3}{5}x + 2 = -\frac{1}{3}x + 2$
Subtract $2$ from both sides:
$\frac{3}{5}x = -\frac{1}{3}x$
Add $\frac{1}{3}x$ to both sides:
$\frac{3}{5}x + \frac{1}{3}x = 0$
Since coefficients are positive, $x$ must be $0$.
$x = 0$

Find $y$:
$y = \frac{3}{5}(0) + 2 = 2$

Solution: $(0, 2)$

10)
Given:
$2x + 3y = -1$
$3x + 4y = -4$

Solve the first equation for $x$:
$2x = -1 - 3y$
$x = \frac{-1 - 3y}{2}$

Substitute into the second equation:
$3(\frac{-1 - 3y}{2}) + 4y = -4$
Multiply by $2$ to clear fraction:
$3(-1 - 3y) + 8y = -8$
$-3 - 9y + 8y = -8$
$-3 - y = -8$
Add $3$ to both sides:
$-y = -5$
$y = 5$

Find $x$:
$x = \frac{-1 - 3(5)}{2} = \frac{-16}{2} = -8$

Solution: $(-8, 5)$

Final Answer:
1) (4, 4)
2) (8, 4)
3) (-3, -8)
4) (-4, -3)
5) (6, 2)
6) (5, -4)
7) (20, -11)
8) (0, -3)
9) (0, 2)
10) (-8, 5)