Two-step equations

1. Solve for \(p\).

\(9(p−4)=−18\)

2. Solve for \(m\).

\(13=2m+5\)

3. Solve for \(y\).

\(6=2(y+2)\)

4. Solve for \(g\).

\(3=\frac{g}{-4}-5\)

5. Solve for \(z\).

\(42=−7(z−3)\)

6. Solve for \(d\).

\(41=12d−7\)

7. Solve for \(q\).

\(3(q−7)=27\)

1. \(p = 2\)

Let's divide and then add to get \(p\) by itself.

\( 9(p-4)=-18 \)

divide each side by \(9\)

\(\begin{array}{r}

\frac{9(p-4)}{9}=\frac{-18}{9} \\

\frac{g(p-4)}{9}=\frac{-18}{9} \\

p-4=\frac{-18}{9}

\end{array}\)

\(p−4 = - 2\)

add \(4\) to each side to get \(p\) by itself

\(\begin{array}{r}

p-4+4=-2+4 \\

p-4+4=-2+4 \\

p=-2+4

\end{array}\)

The answer: \(p = 2\)

Let's check our work!

\(\begin{array}{r}

9(p-4)=-18 \\

9(2-4) \stackrel{?}{=}-18 \\

9(-2) \stackrel{?}{=}-18 \\

-18=-18 \text { Yes! }

\end{array}\)

2. \(m = 4\)

Let's subtract and then divide to get \(m\) by itself.

\(13=2m+5\)

subtract \(5\) from each side

\(\begin{aligned}

&13-5=2 m+5-5 \\

&13-5=2 m+5-\not 5 \\

&13-5=2 m

\end{aligned}\)

\(8 = 2m\)

divide each side by \(2\) to get \(m\) by itself

\(\begin{aligned}

&\frac{8}{2}=\frac{2 m}{2} \\

&\frac{8}{2}=\frac{2 m}{2} \\

&\frac{8}{2}=m

\end{aligned}\)

The answer: \(m = 4\)

Let's check our work!

\(\begin{aligned}

&13=2 m+5 \\

&13 \stackrel{?}{=} 2(4)+5 \\

&13 \stackrel{?}{=} 8+5 \\

&13=13 \quad \text { Yes! }

\end{aligned}\)

3. \(y = 1\)

Let's divide and then subtract to get \(y\) by itself.

\(6=2(y+2)\)

divide each side by \(2\)

\(\begin{aligned}

&\frac{6}{2}=\frac{2(y+2)}{2} \\

&\frac{6}{2}=\frac{2(y+2)}{2} \\

&\frac{6}{2}=y+2

\end{aligned}\)

\(3 = y +2 \)

subtract \(2\) to get \(y\) by itself

\(\begin{aligned}

&3-2=y+2-2 \\

&3-2=y+\not 2-2 \\

&3-2=y

\end{aligned}\)

The answer: \(y = 1\)

Let's check our work!

\(\begin{aligned}

&6=2(y+2) \\

&6 \stackrel{?}{=} 2(1+2) \\

&6 \stackrel{?}{=} 2(3) \\

&6=6 \quad \text { Yes! }

\end{aligned}\)

4. \(g = -32\)

Let's add and then multiply to get \(g\) by itself.

\(3=\frac{g}{-4}-5\)

add \(5\) to each side

\(\begin{aligned}

&3+5=\frac{g}{-4}-5+5 \\

&3+5=\frac{g}{-4}-5+5 \\

&3+5=\frac{g}{-4}

\end{aligned}\)

\(8=\frac{g}{-4}\)

multiply each side by −4 to get \(g\) by itself

\(8 \cdot-4=\frac{g}{-4} \cdot-4\)

\(8 \cdot-4=\frac{g}{-\not 4} \cdot -\not 4\)

\(8 \cdot-4=g\)

Let's check our work!

\(\begin{aligned}

&3=\frac{g}{-4}-5 \\

&3 \stackrel{?}{=} \frac{-32}{-4}-5 \\

&3 \stackrel{?}{=} 8-5 \\

&3=3 \quad \text { Yes! }

\end{aligned}\)

5. \( z = -3 \)

Let's divide and then add to get \(z\) by itself.

\(42=−7(z−3)\)

divide each side by \(−7\)

\(\begin{aligned}

&\frac{42}{-7}=\frac{-7(z-3)}{-7} \\

&\frac{42}{-7}=\frac{7(z-3)}{-7} \\

&\frac{42}{-7}=z-3

\end{aligned}\)

\( -6 = z - 3\)

add \(3\) to each side to get \(z\) by itself

\(\begin{aligned}

&-6+3=z-3+3 \\

&-6+3=z-\not 3+ \not 3 \\

&-6+3=z

\end{aligned}\)

The answer: \(z = -3\)

Let's check our work!

\(\begin{aligned}

&42=-7(z-3) \\

&42 \stackrel{?}{=}-7(-3-3) \\

&42 \stackrel{?}{=}-7(-6) \\

&42=42 \quad \text { Yes! }

\end{aligned}\)

6. \(d = 4\)

Let's add and then divide to get \(d\) by itself.

\(41=12d−7\)

add \(7\) to each side

\(\begin{aligned}

&41+7=12 d-7+7 \\

&41+7=12 d-\not 7+ \not 7 \\

&41+7=12 d

\end{aligned}\)

\(48 = 12d\)

\(\begin{aligned}

&\frac{48}{12}=\frac{12 d}{12} \\

&\frac{48}{12}=\frac{ \not {12} d}{\not {12}} \\

&\frac{48}{12}=d

\end{aligned}\)

The answer: \(d = 4\)

Let's check our work!

\(\begin{aligned}

&41=12 d-7 \\

&41 \stackrel{?}{=} 12(4)-7 \\

&41 \stackrel{?}{=} 48-7 \\

&41=41 \quad \text { Yes! }

\end{aligned}\)

7. \( q = 16\)

Let's divide and then add to get \(q\) by itself.

\(3(q−7)=27\)

divide each side by \(3\)

\(\begin{aligned}

&\frac{3(q-7)}{3}=\frac{27}{3}\\

&\frac{\not 3(q-7)}{\not 3}=\frac{27}{3}\\

&q-7=\frac{27}{3}

\end{aligned}\)

\(q - 7 = 9 \)

add \(7\) to each side to get \(q\) by itself

\(\begin{array}{r}

q-7+7=9+7 \\

q-7+7=9+7 \\

q=9+7

\end{array}\)

The answer: \( q = 16\)

Let's check our work!

\(\begin{aligned}

3(q-7) &=27 \\

3(16-7) & \stackrel{?}{=} 27 \\

3(9) & \stackrel{?}{=} 27 \\

27 &=27 \quad \text { Yes! }

\end{aligned}\)