Now, let’s have you solve a system by substitution.
Which variable should we solve for in order to use substitution and why?
Solve that equation for that variable.
Now, perform the substitution.
And solve for that variable.
| 3(-3 – 2y) – 4y = 16 | substitution | |
| -9 – 6y – 4y = 16 | distribution | |
| -9 – 10y = 16 | combine similar terms | |
| -9 + 9 – 10y = 16 + 9 | add 9 to both sides | |
| -10y = 25 | combine similar terms | |
| -10y ÷ -10 = 25 ÷ -10 | divide both sides by -10 | |
| y = -2.5 | simplify |
Find the value of the other variable.
| x = -3 – 2y | simplest form of the equation | |
| x = -3 – 2(-2.5) | substitution | |
| x = 2 | simplify |
The solution to the system is . . .
If you would like to see how this method would be used in a real-world problem, click Interactive popup. Assistance may be required.
Brad and his brother Andy have 40 video games between them. Brad has five less than twice as many games as Andy. Formulate a system of equations that could be used to determine the number of video games that each brother has.
Define your variables, write a system of equations, and now solve this system using substitution.
Let b = the number of video games that Brad has
Let a = the number of video games that Andy has
If Brad has five less than twice as many as Andy, then b = 2a – 5.
If they have 40 games between them, then b + a = 40.
Consequently, the system modeling this situation is:
b = 2a – 5
b + a = 40
Because the first equation is already solved for b, this system is perfect for solving using substitution.
| b + a = 40 | The second equation | |
| (2a 5) + a = 40 | Substitute the equivalent expression (2a – 5) in for b | |
| 3a – 5 = 40 | Simplify | |
| 3a – 5 + 5 = 40 + 5 | Add 5 to each side of the equation | |
| 3a = 45 | Simplify | |
| 3a / 3 = 45 / 3 | Divide both sides of the equation by 3 | |
| a = 15 | Simplify | |
| b = 2a – 5 | The first equation | |
| b = 2(15) – 5 = 25 | Substitute 15 in for a and simplify |
Brad had 25 video games and Andy had 15 video games.