Systems of Equations - Riverside Math - Free Printable

Here are the solutions for each system of equations. To solve these, we find the point $(x, y)$ where the two lines cross. This point must make both equations true.

1.
$$y = x + 2$$
$$y = -2x + 2$$

* Step 1: Look at the $y$-intercepts (the number added or subtracted at the end). Both equations have $+2$. This means both lines cross the y-axis at $(0, 2)$.
* Step 2: Since they share this point, this is their intersection.
* Check: If $x=0$, $y=2$ for the first equation. If $x=0$, $y=2$ for the second equation.

Solution: $(0, 2)$

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2.
$$y = \frac{2}{3}x$$
$$y = -\frac{4}{3}x + 6$$

* Step 1: Set the equations equal to each other since they both equal $y$.
$$\frac{2}{3}x = -\frac{4}{3}x + 6$$
* Step 2: Add $\frac{4}{3}x$ to both sides to group the $x$ terms.
$$\frac{2}{3}x + \frac{4}{3}x = 6$$
$$\frac{6}{3}x = 6$$
$$2x = 6$$
* Step 3: Divide by 2.
$$x = 3$$
* Step 4: Plug $x=3$ into the first equation to find $y$.
$$y = \frac{2}{3}(3) = 2$$

Solution: $(3, 2)$

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

* Step 1: Set them equal.
$$\frac{1}{5}x + 1 = -\frac{2}{5}x - 4$$
* Step 2: Add $\frac{2}{5}x$ to both sides and subtract $1$ from both sides.
$$\frac{1}{5}x + \frac{2}{5}x = -4 - 1$$
$$\frac{3}{5}x = -5$$
* Step 3: Multiply by $\frac{5}{3}$ to isolate $x$.
$$x = -5 \cdot \frac{5}{3} = -\frac{25}{3}$$ (which is approx $-8.33$)
* Step 4: Find $y$.
$$y = \frac{1}{5}\left(-\frac{25}{3}\right) + 1$$
$$y = -\frac{5}{3} + 1$$
$$y = -\frac{5}{3} + \frac{3}{3} = -\frac{2}{3}$$ (which is approx $-0.67$)

Solution: $\left(-\frac{25}{3}, -\frac{2}{3}\right)$

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4.
$$y = -2x + 2$$
$$y = -4x - 2$$

* Step 1: Set them equal.
$$-2x + 2 = -4x - 2$$
* Step 2: Add $4x$ to both sides and subtract $2$ from both sides.
$$2x = -4$$
* Step 3: Divide by 2.
$$x = -2$$
* Step 4: Find $y$ using the first equation.
$$y = -2(-2) + 2$$
$$y = 4 + 2 = 6$$

Solution: $(-2, 6)$

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5.
$$y = \frac{2}{7}x - 4$$
$$y = -x + 5$$

* Step 1: Set them equal.
$$\frac{2}{7}x - 4 = -x + 5$$
* Step 2: Add $x$ to both sides and add $4$ to both sides.
$$\frac{2}{7}x + 1x = 9$$
*(Note: $1x$ is the same as $\frac{7}{7}x$)*
$$\frac{2}{7}x + \frac{7}{7}x = 9$$
$$\frac{9}{7}x = 9$$
* Step 3: Multiply by $\frac{7}{9}$ to solve for $x$.
$$x = 9 \cdot \frac{7}{9} = 7$$
* Step 4: Find $y$.
$$y = -(7) + 5 = -2$$

Solution: $(7, -2)$

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6.
$$5x - 4y = 16$$
$$-3x + 2y = -12$$

* Step 1: Let's use elimination. Multiply the second equation by $2$ so the $y$ terms cancel out.
$$2 \cdot (-3x + 2y) = 2 \cdot (-12)$$
$$-6x + 4y = -24$$
* Step 2: Add this new equation to the first equation.
$$(5x - 4y) + (-6x + 4y) = 16 + (-24)$$
$$-x = -8$$
$$x = 8$$
* Step 3: Plug $x=8$ into the second original equation.
$$-3(8) + 2y = -12$$
$$-24 + 2y = -12$$
$$2y = 12$$
$$y = 6$$

Solution: $(8, 6)$

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7.
$$y = \frac{1}{3}x - 1$$
$$5x + 3y = 15$$

* Step 1: Substitute the first equation ($y$) into the second equation.
$$5x + 3\left(\frac{1}{3}x - 1\right) = 15$$
* Step 2: Distribute the $3$.
$$5x + 1x - 3 = 15$$
$$6x - 3 = 15$$
* Step 3: Add $3$ to both sides.
$$6x = 18$$
$$x = 3$$
* Step 4: Find $y$.
$$y = \frac{1}{3}(3) - 1$$
$$y = 1 - 1 = 0$$

Solution: $(3, 0)$

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8.
$$-x + 2y = 6$$
$$-3x + 2y = 14$$

* Step 1: Subtract the first equation from the second equation to eliminate $y$.
$$(-3x + 2y) - (-x + 2y) = 14 - 6$$
$$-3x + x = 8$$
$$-2x = 8$$
$$x = -4$$
* Step 2: Plug $x=-4$ into the first equation.
$$-(-4) + 2y = 6$$
$$4 + 2y = 6$$
$$2y = 2$$
$$y = 1$$

Solution: $(-4, 1)$

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Final Answer:
1. Solution: $(0, 2)$
2. Solution: $(3, 2)$
3. Solution: $\left(-\frac{25}{3}, -\frac{2}{3}\right)$
4. Solution: $(-2, 6)$
5. Solution: $(7, -2)$
6. Solution: $(8, 6)$
7. Solution: $(3, 0)$
8. Solution: $(-4, 1)$