Chapter 6, Problem 1RE

Solving a System by Substitution In Exercises 1-10, solve the system by the method of substitution.

$\left\{\begin{array}{c}x+y=2\\ x-y=0\end{array}\right.$

To determine

To calculate: The solution of system of equations, $x+y=2$ and $x-y=0$, by the method of substitution.

Answer to Problem 1RE

The solution of system of equations, $x+y=2$ and $x-y=0$, by the method of substitution is $\left(1,1\right)$.

Explanation of Solution

Calculation:

Consider the provided equations,

$\begin{array}{l}x+y=2\\ x-y=0\end{array}$

Use the method of substitution to solve the above system. Solve any of the equation for one variable in terms of the other.

$y=2-x$ …… (1)

Substitute the expression from equation (1) into the second equation to obtain an equation in one variable as,

$\begin{array}{c}x-y=0\\ x-\left(2-x\right)=0\\ x-2+x=0\\ 2x-2=0\end{array}$

Solve the above equation further, to get

$\begin{array}{c}2x=2\\ x=1\end{array}$

Back-substitute $1$ for $x$ in the above equation to calculate the value of $y$ as,

$\begin{array}{l}y=2-x\\ y=2-1\\ y=1\end{array}$

So, the solution of the system of equations, $x+y=2$ and $x-y=0$, is the ordered pair $\left(1,1\right)$.

Check this solution as follows:

Substitute $1$ for $x$ and $1$ for $y$ in $x+y=2$ as,

$\begin{array}{r}x+y\stackrel{?}{=}2\\ 1+1\stackrel{?}{=}2\\ 2=2\end{array}$

This is true.

Substitute $1$ for $x$ and $1$ for $y$ in $x-y=0$ as,

$\begin{array}{r}x-y\stackrel{?}{=}0\\ 1-1\stackrel{?}{=}0\\ 0=0\end{array}$

This is true.

Point $\left(1,1\right)$ satisfies both the equations in the system.

Therefore, the solution of system of equations, $x+y=2$ and $x-y=0$, by the method of substitution is $\left(1,1\right)$.