Equations with variables on both sides
| Site: | Saylor Academy |
| Course: | GKT101: General Knowledge for Teachers – Math |
| Book: | Equations with variables on both sides |
| Printed by: | Guest user |
| Date: | Tuesday, 20 May 2025, 8:22 AM |
Description
This lecture series shows how you can apply the principle of doing the same thing to both sides of the equation to equations with variables on both sides. Watch the videos and complete the interactive exercise sets.
Table of contents
- Why we do the same thing to both sides: Variable on both sides
- Intro to equations with variables on both sides
- Equations with variables on both sides: 20-7x=6x-6
- Equation with variables on both sides: fractions
- Equation with the variable in the denominator
- Figuring out missing algebraic step
- Equations with variables on both sides - Questions
- Equations with variables on both sides: decimals & fractions - Questions
Why we do the same thing to both sides: Variable on both sides
Source: Khan Academy, https://www.khanacademy.org/math/algebra-home/alg-basic-eq-ineq#alg-variables-on-both-sides
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License.
Intro to equations with variables on both sides
Equations with variables on both sides: 20-7x=6x-6
Equation with variables on both sides: fractions
Equation with the variable in the denominator
Figuring out missing algebraic step
Equations with variables on both sides - Questions
1. Solve for \(e\).
\(9e−7=7e−11\)
2. Solve for \(p\).
\(17−2p=2p+5+2p\)
3. Solve for \(a\).
\(5+14a=9a−5\)
4. Solve for \(m\).
\(−7+4m+10=15−2m\)
Answers
1. \(e=−2 \)
We need to manipulate the equation to get \(e\) by itself.
| \(9e−7=7e−11\) | |
| \(9 e-7-7 e=7 e-11-7 e\) | Subtract \(7e\) from each side. |
| \(2 e-7=-11\) | Combine like terms. |
| \(2 e-7+7=-11+7\) | Add \(7\) to each side. |
| \(2 e=-4\) | Combine like terms. |
| \(\frac{2 e}{2}=\frac{-4}{2}\) | Divide each side by \(2\). |
| \(e=-2\) | Simplify, |
The answer: \(e=−2 \)
Let's check our work!
\(\begin{gathered}
9 e-7=7 e-11 \\
9(-2)-7 \stackrel{?}{=} 7(-2)-11 \\
-18-7 \stackrel{?}{=}-14-11 \\
-25=-25 \quad \text { Yes! }
\end{gathered}\)
2. \(p = 2\)
We need to manipulate the equation to get \(p\) by itself.
| \(17−2p=2p+5+2p\) | |
| \(17-2 p =4 p+5 \) | Combine like terms. |
| \(17-2 p-4 p =4 p+5-4 p \) | Subtract \(4p\) from each side. |
| \(-6 p+17 =5 \) | Combine like terms. |
| \(-6 p+17-17 =5-17 \) | Subtract \(17\) from each side. |
| \(\frac{-6 p}{-6 p} =-12 \) | Combine like terms. |
| \(\frac{-6}{-6} =\frac{-12}{-6} \) | Divide each side by \(-6\) |
| \(p = 2\) | Simplify. |
The answer: \(p = 2\)
Let's check our work!
\(\begin{aligned}
17-2 p &=2 p+5+2 p \\
17-2 p &=4 p+5 \\
17-2(2) & \stackrel{?}{=} 4(2)+5 \\
17-4 & \stackrel{?}{=} 8+5 \\
13 &=13 \quad \text { Yes! }
\end{aligned}\)
3. \(a =-2 \)
We need to manipulate the equation to get \(a\) by itself.
| \(5+14a=9a−5\) | |
| \(5+14 a-9 a =9 a-5-9 a \) | Subtract \(9a\) from each side. |
| \(5+5 a =-5\) | Combine like terms. |
| \(5+5 a-5 =-5-5\) | Subtract \(5\) from each side. |
| \(5 a =-10\) | Combine like terms. |
| \(\frac{5 a}{5} =\frac{-10}{5} \) | Divide each side by \(5\). |
| \(a =-2 \) | Simplify. |
The answer: \(a =-2 \)
Let's check our work!
\(\begin{aligned}
5+14 a &=9 a-5 \\
5+14(-2) & \stackrel{?}{=} 9(-2)-5 \\
5+(-28) & \stackrel{?}{=}-18-5 \\
-23 &=-23 \quad \text { Yes! }
\end{aligned}\)
4. \(m = 2\)
We need to manipulate the equation to get \(m\) by itself.
| \(−7+4m+10=15−2m \) | |
| \( 4 m+3 =15-2 m \) | Combine like terms. |
| \( 3+4 m+2 m =15-2 m+2 m \) | Add \(2m\) to each side. |
| \( 6 m+3 =15 \) | Combine like terms. |
| \( 6 m+3-3 =15-3 \) | Subtract \(3\) from each side. |
| \( 6 m =12 \) | Combine like terms. |
| \( \frac{6 m}{6} =\frac{12}{6} \) | Divide each side by \(6\). |
| \( m = 2\) | Simplify. |
The answer: \(m = 2\)
Let's check our work!
\(\begin{aligned}
-7+4 m+10 &=15-2 m \\
3+4 m &=15-2 m \\
3+4(2) & \stackrel{?}{=} 15-2(2) \\
3+8 & \stackrel{?}{=} 15-4 \\
11 &=11 \quad \text { Yes! }
\end{aligned}\)
Equations with variables on both sides: decimals & fractions - Questions
1. Solve for \(k\).
\(4.5+1.5k=18−3k\)
2. Solve for \(s\).
\(2-2 s=\frac{3}{4} s+13\)
3. Solve for \(g\).
\(9+3.5g=11−0.5g\)
4. Solve for \(p\).
\(16-3 p=\frac{2}{3} p+5\)
Answers
1. \(k = 3\)
We need to manipulate the equation to get \(k\) by itself.
| \(4.5+1.5k=18−3k\) | |
| \( 4.5+1.5 k+3 k =18-3 k+3 k \) | Add \(3k\) to each side. |
| \( 4.5 k+4.5 =18 \) | Combine like terms. |
| \( 4.5 k+4.5-4.5 =18-4.5 \) | Subtract \(4.5\) from each side. |
| \( 4.5 k =13.5 \) | Combine like terms. |
| \( \frac{4.5 k}{4.5} =\frac{13.5}{4.5} \) | Divide each side by \(4.5\). |
| \( k =3 \) | Simplify. |
The answer: \(k = 3\)
Let's check our work!
\(\begin{aligned}
4.5+1.5 k &=18-3 k \\
4.5+1.5(3) & \stackrel{?}{=} 18-3(3) \\
4.5+4.5 & \stackrel{?}{=} 18-9 \\
9 &=9 \quad \text { Yes! }
\end{aligned}\)
2.
We need to manipulate the equation to get \(s\) by itself.
| \(2-2 s=\frac{3}{4} s+13\) | |
| \( 2-2 s-\frac{3}{4} s =\frac{3}{4} s+13-\frac{3}{4} s\) | Subtract \(\frac{3}{4}\) from each side. |
| \( -\frac{11}{4} s+2 =13 \) | Combine like terms. |
| \( -\frac{11}{4} s+2-2 =13-2 \) | Subtract \(2\) from each side. |
| \( -\frac{11}{4} s =11 \) | Combine like terms. |
| \( s \cdot\left(-\frac{4}{11}\right) =11 \cdot\left(-\frac{4}{11}\right)\) | Multiply each side by \(-\frac{4}{11}\). |
| \( s =-\frac{44}{11} \) | |
| \( s =-4 \) | Simplify. |
The answer: \(s = -4\)
Let's check our work!
\(\begin{aligned}
2-2 s &=\frac{3}{4} s+13 \\
2-2(-4) & \stackrel{?}{=} \frac{3}{4}(-4)+13 \\
2+8 & \stackrel{?}{=}-\frac{12}{4}+13 \\
10 & \stackrel{?}{=}-3+13 \\
10 &=10 \quad \text { Yes! }
\end{aligned}\)
3. \(g = 0.5\)
We need to manipulate the equation to get \(g\) by itself.
| \(9+3.5g=11−0.5g\) | |
| \( 9+3.5 g+0.5 g =11-0.5 g+0.5 g \) | Add \(0.5g\) to each side. |
| \( 9+4 g =11 \) | Combine like terms. |
| \( 4 g+9-9 =11-9 \) | Subtract \(9\) from each side. |
| \( 4 g =2 \) | Combine like terms. |
| \( \frac{4 g}{4} =\frac{2}{4} \) | Divide each side by \(4\). |
| \( g =0.5 \) | Simplify. |
The answer: \(g = 0.5\)
Let's check our work!
\(\begin{aligned}
9+3.5 g &=11-0.5 g \\
9+3.5(0.5) & \stackrel{?}{=} 11-0.5(0.5) \\
9+1.75 & \stackrel{?}{=} 11-0.25 \\
10.75 &=10.75 \quad \text { Yes! }
\end{aligned}\)
4. \(p = 3\)
We need to manipulate the equation to get \(p\) by itself.
| \(16-3 p=\frac{2}{3} p+5\) | |
| \( 16-3 p-\frac{2}{3} p =\frac{2}{3} p+5-\frac{2}{3} p \) | Subtract \(\frac{2}{3}p\) from each side. |
| \( -\frac{11}{3} p+16 =5 \) | Combine like terms. |
| \( -\frac{11}{3} p+16-16 =5-16 \) | Subtract \(16\) from each side. |
| \( -\frac{11}{3} p =-11 \) | Combine like terms. |
| \( -\frac{11}{3} p \cdot\left(-\frac{3}{11}\right) =-11 \cdot\left(-\frac{3}{11}\right) \) | Multiply each side by \(-\frac{3}{11}\) |
| \( p =\frac{33}{11}\) | |
| \( p =3 \) | Simplify. |
The answer: \(p = 3\)
Let's check our work!
\(\begin{aligned}
16-3 p &=\frac{2}{3} p+5 \\
16-3(3) & \stackrel{?}{=} \frac{2}{3}(3)+5 \\
16-9 & \stackrel{?}{=} 2+5 \\
7 &=7 \quad \text { Yes! }
\end{aligned}\)