Question

Solve the system of equation$\left\{\begin{array}{l}-2x+y=10\\ -x+2y=16\end{array}\right.$by substitution

Short answer

The solution is$\left(-\frac{4}{3},\frac{22}{3}\right)$.

Step-by-step solution

Step 1. Given information 

The given equations are:

$-2x+y=10\cdots \cdots \left(1\right)$

$-x+2y=16\cdots \cdots \left(2\right)$

Step 2. Substitute value of y from equation 1 in equation 2

From equation $\left(1\right)$,$y=10+2x$

Put in equation $\left(2\right)$

$\Rightarrow -x+2\left(10+2x\right)=16$

Step 3. Solve for x

$\Rightarrow -x+20+4x=16$

$\Rightarrow 3x=16-20$

$\Rightarrow 3x=-4$

$\Rightarrow x=-\frac{4}{3}$

Step 4. Substitute value of x from step 3 in equation 2

$\Rightarrow -\left(-\frac{4}{3}\right)+2y=16$

$\Rightarrow 2y=16-\frac{4}{3}$

$\Rightarrow 2y=\frac{48-4}{3}$

$\Rightarrow 2y=\frac{44}{3}$

$\Rightarrow y=\frac{22}{3}$

Step 5. Checking solution 

Now we have$x=-\frac{4}{3}$and$y=\frac{22}{3}$

Put these values in equation $\left(1\right)$

$\Rightarrow -2\left(-\frac{4}{3}\right)+\frac{22}{3}=10$

$\Rightarrow 10=10$

Also put these values in equation $\left(2\right)$

$\Rightarrow -\left(-\frac{4}{3}\right)+2\left(\frac{22}{3}\right)=16$

$\Rightarrow 16=16$