Algebra: Solve Linear Equation Practice 2
In this section, you will work through 6 carefully selected questions. Each question is designed to help you apply and master the balancing equation method when solving algebraic linear equations. Focus on showing clear working and applying the technique accurately at every step.
Related Lessons:
Practice 2 - Question 1
Solve the equation \(3x+7\left(x+2\right)=4\left(5-3x\right)+7\)
Solution:
\[
\begin{align*}
3x+7\left(x+2\right)&=4\left(5-3x\right)+7\\
3x+7x+14&=20-12x+7\\
10x+14&=27-12x\\
10x+12x&=27-14\\
22x&=13\\
x&=\frac{13}{22}\\
\end{align*}
\]
Practice 2 - Question 2
Solve the equation \(\frac{b}{2}+\frac{b+5}{3}=2\)
Solution:
\[
\begin{align*}
\frac{b}{2}+\frac{b+5}{3}&=2\ \ \ \ \ \ \ \ \ \ \ \ \ \times6\\
3b+2\left(b+5\right)&=12\\
3b+2b+10&=12\\
5b+10&=12\\
5b&=12-10\\
b&=\frac{2}{5}\\
\end{align*}
\]
Practice 2 - Question 3
Solve the equation \(3\left(x+5\right)=8x-7\)
Solution:
\[
\begin{align*}
3\left(x+5\right)&=8x-7\\
3x+15&=8x-7\\
3x-8x&=-7-15\\
-5x&=-22\\
x&=\frac{-22}{-5}\\
x&=4\frac{2}{5}\\
\end{align*}
\]
Practice 2 - Question 4
Solve the equation \(\frac{2x-1}{5}=\frac{x+3}{8}\)
Solution:
\[
\begin{align*}
\frac{2x-1}{5}&=\frac{x+3}{8}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \times40\\
\left(2x-1\right)&=5\left(x+3\right)\\
16x-8&=5x+15\\
16x-5x&=15+8\\
11x&=23\\
x&=\frac{23}{11}\\
x&=2\frac{1}{11}\\
\end{align*}
\]
Practice 2 - Question 5
Solve the equation \(8a-4=5a+20\)
Solution:
\[
\begin{align*}
8a-4&=5a+20\\
8a-5a&=20+4\\
3a&=24\\
a&=\frac{24}{3}\\
a&=8\\
\end{align*}
\]