When solving a using the , there are three steps.
- Isolate a in one of the equations.
- Substitute the expression for that variable into the other equation and solve.
- Substitute this solution into one of the equations and solve for the value of the other variable.
For this exercise,
y is already isolated in both equations, so we can skip straight to solving! Since the expression equal to
y in (II) is simpler, let's use that for our initial substitution.
{y=2x+3y=5x(I)(II)
{5x=2x+3y=5x
{5x−2x=2x+3−2xy=5x
{3x=3y=5x
{33x=33y=5x
{x=1y=5x
Great! Now, to find the value of
y, we need to substitute
x=1 into either one of the equations in the given system. Let's use the second equation.
{x=1y=5x
{x=1y=5(1)
{x=1y=5
The solution, or , to this system of equations is the point
(1,5).
Checking Our Answer
To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.
{y=2x+3y=5x(I)(II)
{5=?2(1)+35=?5(1)
(I), (II): Identity Property of Multiplication
{5=?2+35=5
{5=5✓5=5✓
Because both equations are true statements, we know that our solution is correct.