Question

How do you find the ordered pair that is a solution to the system of equations $x+y=10$ and $x-y=8$?

Answer 1

Find the solution to the system of equations:

Given two expressions are $x+y=10$ and $x-y=8$.

Need to find all solutions $\left(x,y\right)$ that satisfies the equations $x+y=10$ and $x-y=8$.

Write system of equations as,

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

$x-y=8\cdots \left(2\right)$

Add equation $\left(1\right)$ and equation $\left(2\right)$.

$$\begin{gathered} x+y+\left(x-y\right)=10+8 \\ \Rightarrow x+y+x-y=18 \\ \Rightarrow 2x=18 \\ \Rightarrow x=\frac{18}{2}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2\right] \\ \therefore x=9 \end{gathered}$$

Substitute $x=9$ into equation $\left(2\right)$.

$$\begin{gathered} 9+y=10 \\ \Rightarrow 9+y-9=10-9\left[\mathrm{Subtracting}\mathrm{both}\mathrm{sides}\mathrm{by}9\right] \\ \therefore y=1 \end{gathered}$$

Therefore, $\left(x,y\right)=\left(9,1\right)$.

Hence, the solution to the system of equations $x+y=10$ and $x-y=8$ is $\left(9,1\right)$.