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Q. 102

Expert-verifiedFound in: Page 611

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.

The given equations are

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

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*.

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

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.

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