Systems of Equations - Solving Two Variable Systems of Equations by Substitution - Free Printable

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

1)

System:
$y = \frac{7}{4}x - 3$
$y = 4$

Step 1: Since we know $y = 4$, substitute $4$ for $y$ in the first equation.
$$4 = \frac{7}{4}x - 3$$

Step 2: Add $3$ to both sides to isolate the term with $x$.
$$7 = \frac{7}{4}x$$

Step 3: Multiply both sides by $\frac{4}{7}$ (the reciprocal of $\frac{7}{4}$) to solve for $x$.
$$x = 7 \cdot \frac{4}{7}$$
$$x = 4$$

Solution: $(4, 4)$

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2)

System:
$y = -\frac{1}{4}x + 6$
$y = 4$

Step 1: Substitute $4$ for $y$ in the first equation.
$$4 = -\frac{1}{4}x + 6$$

Step 2: Subtract $6$ from both sides.
$$-2 = -\frac{1}{4}x$$

Step 3: Multiply both sides by $-4$ to solve for $x$.
$$x = (-2) \cdot (-4)$$
$$x = 8$$

Solution: $(8, 4)$

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3)

System:
$6x - 5y = 22$
$y = -8$

Step 1: Substitute $-8$ for $y$ in the first equation.
$$6x - 5(-8) = 22$$

Step 2: Simplify the multiplication ($-5 \cdot -8 = 40$).
$$6x + 40 = 22$$

Step 3: Subtract $40$ from both sides.
$$6x = 22 - 40$$
$$6x = -18$$

Step 4: Divide by $6$.
$$x = -3$$

Solution: $(-3, -8)$

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4)

System:
$4x - 7y = 5$
$9x - 7y = -15$

Step 1: Solve the first equation for $7y$ (or isolate a variable). Let's rearrange the first equation:
$$4x - 5 = 7y$$

Step 2: Substitute $(4x - 5)$ for $7y$ in the second equation. Note that the second equation has $-7y$, so we subtract $(4x - 5)$.
$$9x - (4x - 5) = -15$$

Step 3: Distribute the negative sign and simplify.
$$9x - 4x + 5 = -15$$
$$5x + 5 = -15$$

Step 4: Subtract $5$ from both sides.
$$5x = -20$$
$$x = -4$$

Step 5: Substitute $x = -4$ back into the expression for $7y$:
$$7y = 4(-4) - 5$$
$$7y = -16 - 5$$
$$7y = -21$$
$$y = -3$$

Solution: $(-4, -3)$

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5)

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

Step 1: Isolate $x$ in the second equation because the numbers are smaller.
$$3x = 22 - 2y$$
$$x = \frac{22 - 2y}{3}$$

Step 2: Substitute this expression for $x$ into the first equation.
$$-5\left(\frac{22 - 2y}{3}\right) + 9y = -12$$

Step 3: To clear the fraction, multiply the entire equation by $3$.
$$-5(22 - 2y) + 27y = -36$$

Step 4: Distribute the $-5$.
$$-110 + 10y + 27y = -36$$
$$-110 + 37y = -36$$

Step 5: Add $110$ to both sides.
$$37y = 74$$
$$y = 2$$

Step 6: Substitute $y = 2$ back into the isolated $x$ equation.
$$x = \frac{22 - 2(2)}{3}$$
$$x = \frac{22 - 4}{3}$$
$$x = \frac{18}{3}$$
$$x = 6$$

Solution: $(6, 2)$

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6)

System:
$y = -\frac{2}{5}x - 2$
$y = -4$

Step 1: Substitute $-4$ for $y$ in the first equation.
$$-4 = -\frac{2}{5}x - 2$$

Step 2: Add $2$ to both sides.
$$-2 = -\frac{2}{5}x$$

Step 3: Multiply both sides by $-\frac{5}{2}$ to solve for $x$.
$$x = (-2) \cdot \left(-\frac{5}{2}\right)$$
$$x = 5$$

Solution: $(5, -4)$

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7)

System:
$y = -\frac{3}{4}x + 4$
$y = -\frac{1}{2}x - 1$

Step 1: Set the two expressions for $y$ equal to each other.
$$-\frac{3}{4}x + 4 = -\frac{1}{2}x - 1$$

Step 2: To remove fractions, multiply every term by $4$ (the least common denominator).
$$-3x + 16 = -2x - 4$$

Step 3: Add $3x$ to both sides to make $x$ positive.
$$16 = x - 4$$

Step 4: Add $4$ to both sides.
$$20 = x$$

Step 5: Substitute $x = 20$ into either original equation to find $y$. Let's use the second one.
$$y = -\frac{1}{2}(20) - 1$$
$$y = -10 - 1$$
$$y = -11$$

Solution: $(20, -11)$

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8)

System:
$y = -\frac{4}{9}x - 3$
$y = -\frac{7}{5}x - 3$

Step 1: Set the equations equal to each other.
$$-\frac{4}{9}x - 3 = -\frac{7}{5}x - 3$$

Step 2: Add $3$ to both sides. The constants cancel out.
$$-\frac{4}{9}x = -\frac{7}{5}x$$

Step 3: For these two terms to be equal, $x$ must be $0$ (since the coefficients $-\frac{4}{9}$ and $-\frac{7}{5}$ are different).
$$x = 0$$

Step 4: Substitute $x = 0$ into either equation to find $y$.
$$y = -\frac{4}{9}(0) - 3$$
$$y = -3$$

Solution: $(0, -3)$

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9)

System:
$y = \frac{3}{5}x + 2$
$y = -\frac{1}{3}x + 2$

Step 1: Set the equations equal to each other.
$$\frac{3}{5}x + 2 = -\frac{1}{3}x + 2$$

Step 2: Subtract $2$ from both sides.
$$\frac{3}{5}x = -\frac{1}{3}x$$

Step 3: Similar to problem 8, since the coefficients are different, the only way they are equal is if $x$ is zero.
$$x = 0$$

Step 4: Substitute $x = 0$ into either equation.
$$y = \frac{3}{5}(0) + 2$$
$$y = 2$$

Solution: $(0, 2)$

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10)

System:
$2x + 3y = -1$
$3x + 4y = -4$

Step 1: Isolate $x$ in the first equation.
$$2x = -1 - 3y$$
$$x = \frac{-1 - 3y}{2}$$

Step 2: Substitute this into the second equation.
$$3\left(\frac{-1 - 3y}{2}\right) + 4y = -4$$

Step 3: Multiply the entire equation by $2$ to clear the fraction.
$$3(-1 - 3y) + 8y = -8$$

Step 4: Distribute the $3$.
$$-3 - 9y + 8y = -8$$
$$-3 - y = -8$$

Step 5: Add $3$ to both sides.
$$-y = -5$$
$$y = 5$$

Step 6: Substitute $y = 5$ back into the isolated $x$ equation.
$$x = \frac{-1 - 3(5)}{2}$$
$$x = \frac{-1 - 15}{2}$$
$$x = \frac{-16}{2}$$
$$x = -8$$

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)