Systems of Equations—Solving with Substitution Study Guide
Introduction to Systems of Equations—Solving with Substitution
In mathematics, you don't understand things. You just get used to them.
— Johann Von Neumann (1903–1957) Hungarian–American Mathematician
In this lesson, you'll learn how to solve systems of equations with two or three variables using substitution.
We learned how to solve an equation with one variable in Lesson 9. In that same lesson, we saw that we could not find the values of two variables with just one equation. The best we could do was to solve for one variable in terms of the other. For example, if 3x + y = 7, we could subtract 3x from both sides of the equation to solve for y in terms of x:
- 3x – 3x + y = 7 – 3x
- y = 7 – 3x
Once we had y in terms of x, there was nothing more we could do. In order to find the values of x and y, we need a second, related equation. A group of two or more equations for which the common variables in each equation have the same values is called a system of equations.
The equation 3x + y = 7 and the equation x – y = 13 form a system of equations. The values of x and y in the first equation are also the values of x and y in the second equation. Now that we have two equations, we can find the values of both variables.
We have a couple of ways to solve a system of equations. In this lesson, we will learn how to solve them using substitution. In the next lesson, we will learn how to solve them using elimination.
When we solve a system of equations using substitution, we begin by writing one variable in terms of the other. In this example, we wrote y in terms of x: y = 7 – 3x. Next, we replace that variable in the other equation with its expression. In this example, we replace the y in the second equation, x – y = 13, with 7 – 3x, because that is the value of y. The second equation is now x – (7 – 3x) = 13. Now we have a single equation with a single variable. Solve for the value of the variable:
- x – (7 – 3x) = 13
- x – 7 + 3x = 13
- 4x – 7 = 13
- 4x – 7 + 7 = 13 + 7
- 4x = 20
- x = 5
We now have the value of one variable, x. We can replace x with its value in either equation to find the value of y. Let's use the second equation:
- x – y = 13
- 5 – y = 13
- –y = 8
- y = –8
The solution to this system of equations is x = 5, y = –8. Substitute the value of each variable into both equations to check the answer:
|3x + y = 7||x – y = 13|
|3(5) + (–8) = 7||5 – (–8) = 13|
|15 – 8 = 7||5 + 8 = 13|
|7 = 7||13 = 13|
If a variable in one equation of a system of equations does not have a coefficient, begin by solving for that variable in terms of the other variable. By avoiding division, you may avoid working with fractions.
- 3y = 12x + 3
- 5x + y = 10
Start by writing one variable in terms of the other. We could write x in terms of y using either equation, or we could write y in terms of x using either equation. Because y has no coefficient in the second equation, we will start by writing y in terms of x using that equation:
- 5x + y = 10
- y = 10 – 5x
Next, replace y in the first equation with the expression that is equal to y, 10 – 5x:
- 3(10 – 5x) = 12x + 3
Solve for the value of x :
- 3(10 – 5x ) = 12x + 3
- 30 – 15x = 12x + 3
- 30 – 15x + 15x = 12x + 15x + 3
- 30 = 27x + 3
- 30 – 3 = 27x + 3 – 3
- 27 = 27x
- 1 = x
We have the value of one variable, x. Replace x with its value in either equation and solve for the value of y:
- 5(1) + y = 10
- 5 + y = 10
- 5 – 5 + y = 10 – 5
- y = 10 – 5
- y = 5
The solution to this system of equations is x = 1, y = 5.
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