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Q. 102

Expert-verified
Found in: Page 611

### Elementary Algebra

Book edition 2nd
Author(s) Lynn Marecek, MaryAnne Anthony-Smith
Pages 1240 pages
ISBN 9780998625713

# In the following exercises, solve the systems of equations by substitution. $15x+4y=6\phantom{\rule{0ex}{0ex}}-30x-8y=-12$

The equations have infinitely many solutions.

See the step by step solution

## Step 1. Given Information

The given equations are

$15x+4y=6\phantom{\rule{0ex}{0ex}}-30x-8y=-12$

## Step 2. Calculation

Solve the second equation for variable y.

$-30x-8y=-12\phantom{\rule{0ex}{0ex}}-8y=-12+30x\phantom{\rule{0ex}{0ex}}y=\frac{-12+30x}{-8}\phantom{\rule{0ex}{0ex}}y=\frac{3}{2}-\frac{15}{4}x$

Substitute the obtained value of variable y in the first equation to find the value of variable x.

$15x+4y=6\phantom{\rule{0ex}{0ex}}15x+4\left(\frac{3}{2}-\frac{15}{4}x\right)=6\phantom{\rule{0ex}{0ex}}15x+6-15x=6\phantom{\rule{0ex}{0ex}}6=6$

Since $6=6$ is a true statement, the system is consistent. The equations are dependent. The graph of these two equations would give the same line. Thus the system of the linear equation has infinitely many solutions.