When solving a using , there are three steps.
- Isolate a in one of the .
- 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.
We have been given the following system of equations.
3x-5y=13 & (I) x+4y=10 & (II)
Let's start by isolating the x variable in Equation (II).
x+4y=10 ⇔ x=10-4y
Now, we can substitute (II) into (I) and solve for y.
3x-5y=13 & (I) x=10-4y & (II)
3( 10-4y)-5y=13 x=10-4y
30-12y-5y=13 x=10-4y
30-17y=13 x=10-4y
-17y=-17 x=10-4y
y=1 x=10-4y
Now, to find the value of x, we need to substitute y=1 into either one of the equations in the given system. Let's use the second equation.
y=1 & (I) x=10-4y & (II)
y=1 x=10-4( 1)
y=1 x=10-4
y=1 x=6
The solution, or , to this system of equations is the point (6,1).
Checking Our Solution
To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.
3x-5y=13 & (I) x=10-4y & (II)
3(6)-5(1)=13 6=10-4(1)
18-5(1)=13 6=10-4(1)
(I), (II): Identity Property of Multiplication
18-5=13 6=10-4
13=13 6=6
Because both equations are true statements, we know that our solution is correct.
