Two-step equations
Two-step equations - Questions
Answers
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!
\(\begin{aligned}
3(q-7) &=27 \\
3(16-7) & \stackrel{?}{=} 27 \\
3(9) & \stackrel{?}{=} 27 \\
27 &=27 \quad \text { Yes! }
\end{aligned}\)