Question

1. solve each system of equations without graphing.

a. {4d + 7e = 68
{-4d - 6e = -72
b. {x + y = 1
{x - y = 3}

Explanation:

Step1: Solve the first - system of equations for \(d\) and \(e\)

We have the system \(

$$\begin{cases}4d + 7e=68\\- 4d+6e=-72\end{cases}$$

\). Add the two equations together to eliminate \(d\).
\((4d + 7e)+(-4d + 6e)=68+( - 72)\)
\(4d-4d+7e + 6e=68 - 72\)
\(13e=-4\)
\(e =-\frac{4}{13}\)

Step2: Substitute \(e\) into the first equation to find \(d\)

Substitute \(e =-\frac{4}{13}\) into \(4d + 7e=68\).
\(4d+7\times(-\frac{4}{13})=68\)
\(4d-\frac{28}{13}=68\)
\(4d=68+\frac{28}{13}\)
\(4d=\frac{68\times13 + 28}{13}=\frac{884+28}{13}=\frac{912}{13}\)
\(d=\frac{912}{13\times4}=\frac{228}{13}\)

Step3: Solve the second - system of equations for \(x\) and \(y\)

We have the system \(

$$\begin{cases}x + y=1\\x - y=3\end{cases}$$

\). Add the two equations together to eliminate \(y\).
\((x + y)+(x - y)=1 + 3\)
\(x+x+y - y=4\)
\(2x=4\)
\(x = 2\)

Step4: Substitute \(x\) into the first equation to find \(y\)

Substitute \(x = 2\) into \(x + y=1\).
\(2+y=1\)
\(y=1 - 2=-1\)

Answer:

a. \(d=\frac{228}{13},e =-\frac{4}{13}\)
b. \(x = 2,y=-1\)