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Find the solution to the following system.

$\left \{ \begin{array}{cc} {y}={4}{x}+{14}\\{2}{x}-{7}{y}={-20} \end{array} \right.$

A.

${x}={-3}, \;{y}={2}$ A.

${x}={-3}, \;{y}={2}$
B.

${x}={2}, \;{y}={-3}$
C.

${x}={3}, \;{y}={2}$
D.

${x}={-3}, \;{y}={-2}$
E.

${x}={-2}, \;{y}={3}$

1. We are given that

$\;{y}={4}{x}+{14}.\;$ We can substitute this expression into the second equation and solve for

$\;{x}.$

$\;\;\;{2}{x}-{7}({4}{x}+{14})={-20}\;$

$\;\;\; {2}{x}-{28}{x}-{98}={-20}\;$

$\;\;\; {-26}{x} = {78}\;\;\;\;\; \therefore {x}={-3}\;$
Now, substitute this value for

$\;{x}\;$ into the first equation of

$\;{y}={4}{x}+{14}={4}\cdot{-3}+{14}={2}\;$
Therefore, the solution to the system of equations is

$\;{x}={-3}, \;{y}={2}$
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