Solve the following systems using the ELIMINATION METHOD. y=4x-5 y=3x+7

1 answer:

7 0

$\left\{\begin{array}{ccc}y=4x-5\\y=3x+7&|\text{change the signs}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}y=4x-5\\-y=-3x-7\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad0=x-12\qquad\text{add 12 to both sides}\\\qquad\boxed{x=12}\\\\\text{Substitute the value of x to the first equation}\\\\y=4(12)-5\\y=48-5\\\boxed{y=43}\\\\Answer:\ x=12\ and\ y=43.$

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$\bf ~~~~~~ \textit{Simple Interest Earned Amount}
\\\\
A=P(1+rt)\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &\$2555\\
P=\textit{original amount deposited}\\
r=rate\to 5.5\%\to \frac{5.5}{100}\to &0.055\\
t=years\to &1
\end{cases}
\\\\\\
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\\\\\\
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