Worksheet for solving systems of equations using substitution, featuring ten algebra problems.

Here are the step-by-step solutions for each problem on the worksheet.

1) $y = -3x - 3$ and $y = -3$
* Since we know $y = -3$, substitute $-3$ for $y$ in the first equation:
$-3 = -3x - 3$
* Add $3$ to both sides:
$0 = -3x$
* Divide by $-3$:
$x = 0$
* Solution: $(0, -3)$

2) $3x + 2y = -15$ and $2x = -18$
* Solve the second equation for $x$ first:
$x = -18 / 2$
$x = -9$
* Substitute $x = -9$ into the first equation:
$3(-9) + 2y = -15$
$-27 + 2y = -15$
* Add $27$ to both sides:
$2y = 12$
* Divide by $2$:
$y = 6$
* Solution: $(-9, 6)$

3) $y = -\frac{13}{4}x + 7$ and $y = \frac{3}{4}x - 9$
* Set the two equations equal to each other since they both equal $y$:
$-\frac{13}{4}x + 7 = \frac{3}{4}x - 9$
* Add $\frac{13}{4}x$ to both sides to group the $x$ terms:
$7 = \frac{16}{4}x - 9$
* Simplify $\frac{16}{4}$ to $4$:
$7 = 4x - 9$
* Add $9$ to both sides:
$16 = 4x$
* Divide by $4$:
$x = 4$
* Plug $x = 4$ back into the second equation to find $y$:
$y = \frac{3}{4}(4) - 9$
$y = 3 - 9$
$y = -6$
* Solution: $(4, -6)$

4) $-5x + 9y = -12$ and $3x + 2y = 22$
* Solve the second equation for $x$ (it's easier because the coefficient is smaller):
$3x = 22 - 2y$
$x = \frac{22 - 2y}{3}$
* Substitute this into the first equation:
$-5(\frac{22 - 2y}{3}) + 9y = -12$
* Multiply the whole equation by $3$ to clear the fraction:
$-5(22 - 2y) + 27y = -36$
$-110 + 10y + 27y = -36$
$-110 + 37y = -36$
* Add $110$ to both sides:
$37y = 74$
* Divide by $37$:
$y = 2$
* Plug $y = 2$ back into the equation for $x$:
$x = \frac{22 - 2(2)}{3}$
$x = \frac{18}{3}$
$x = 6$
* Solution: $(6, 2)$

5) $y = -\frac{2}{5}x - 2$ and $y = -4$
* Substitute $-4$ for $y$ in the first equation:
$-4 = -\frac{2}{5}x - 2$
* Add $2$ to both sides:
$-2 = -\frac{2}{5}x$
* Multiply both sides by $-\frac{5}{2}$ (the reciprocal) to isolate $x$:
$x = -2 \cdot (-\frac{5}{2})$
$x = 5$
* Solution: $(5, -4)$

6) $y = \frac{7}{4}x - 3$ and $y = 4$
* 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 $\frac{4}{7}$:
$x = 7 \cdot \frac{4}{7}$
$x = 4$
* Solution: $(4, 4)$

7) $4x + 8y = 20$ and $-4x + 2y = -30$
* Solve the first equation for $4x$ (or simplify it first). Let's simplify the first equation by dividing by 4:
$x + 2y = 5 \rightarrow x = 5 - 2y$
* Substitute $(5 - 2y)$ for $x$ in the second equation:
$-4(5 - 2y) + 2y = -30$
$-20 + 8y + 2y = -30$
$-20 + 10y = -30$
* Add $20$ to both sides:
$10y = -10$
* Divide by $10$:
$y = -1$
* Plug $y = -1$ back into the simplified first equation:
$x = 5 - 2(-1)$
$x = 5 + 2$
$x = 7$
* Solution: $(7, -1)$

8) $-4x - 15y = -17$ and $-x + 5y = -13$
* Solve the second equation for $x$. It is easy to isolate:
$-x = -13 - 5y$
$x = 13 + 5y$
* Substitute $(13 + 5y)$ for $x$ in the first equation:
$-4(13 + 5y) - 15y = -17$
$-52 - 20y - 15y = -17$
$-52 - 35y = -17$
* Add $52$ to both sides:
$-35y = 35$
* Divide by $-35$:
$y = -1$
* Plug $y = -1$ back into the equation for $x$:
$x = 13 + 5(-1)$
$x = 13 - 5$
$x = 8$
* Solution: $(8, -1)$

9) $6x - 5y = 22$ and $y = -8$
* 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: $(-3, -8)$

10) $y = \frac{1}{2}x + 5$ and $y = -\frac{5}{2}x - 1$
* Set the equations equal to each other:
$\frac{1}{2}x + 5 = -\frac{5}{2}x - 1$
* Add $\frac{5}{2}x$ to both sides:
$\frac{6}{2}x + 5 = -1$
* Simplify $\frac{6}{2}$ to $3$:
$3x + 5 = -1$
* Subtract $5$ from both sides:
$3x = -6$
* Divide by $3$:
$x = -2$
* Plug $x = -2$ into the first equation to find $y$:
$y = \frac{1}{2}(-2) + 5$
$y = -1 + 5$
$y = 4$
* Solution: $(-2, 4)$

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