How to Solve Systems of Equations by Substitution
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- Solving Systems of Equations by Substitution
Overview of Solving Systems of Equations by Substitution:
The Substitution Method
Substitution is the quickest method of solving a system of two equations in two variables. The method can also be used to find the solution of a system of three or more equations in three or more variables, but it takes longer.
In this lesson, we are going to talk you through solving a system of 2 linear equations in 2 variables.
Solving Systems of Equations by Substitution
The substitution method involves three steps. They are:
✯ Rearrange an equation to isolate one of the variables on one side.
✯ Substitute the expression so obtained into the other equation to solve for the other variable.
✯ Plug the value back into one of the equations to solve for the variable initially isolated.
Let's understand the steps with this linear system.
y = x + 6
5x + 2y = 19
In the above system, the variable y is isolated. So, the step 1 is already accomplished. Now, let’s substitute the expression for y in the second equation.
5x + 2(x + 6) = 19
Applying the distributive property, we have:
5x + 2x + 12 = 19
Combining like terms, we get:
7x + 12 = 19
Subtracting 12 from both sides, we have:
7x + 12 –12 = 19 –12
⇨ 7x = 7
Dividing both sides by 7, we get:
7x7 = 77
⇨ x = 1
Now the last step! We need to plug this value of x back in one of the equations. Let’s plug it in y = x + 6.
⇨ y = 1 + 6
⇨ y = 7
Thus, the solution of the linear system is (1, 7).
Let’s take a look at another example.
Example
Solve 4x – 6y = –16 and 8x + 2y = 24 using the substitution method.
Let’s choose the second equation this time. Solving the second equation for y, we get:
8x + 2y = 24
2y = –8x + 24
⇨ y = –4x + 12
Now substituting y in the first equation, we have:
4x – 6(–4x + 12) = –16
4x + 24x – 72 = –16
28x = –16 + 72
28x = 56
⇨ x = 2
We can put this x-value back into any of the given equations and solve for y.
But,do you remember?
We have rearranged the equation 2 and have an expression for y all set!
⇨ y = –4x + 12
Now, all we need to do is just substitute the value of x into this.
y = –4x + 12
y = –4(2) + 12
y = –8 + 12
⇨ y = 4
Thus, the solution is (x, y) = (2, 4).
Check Your Solution!
You can cross-check if your solution is right by substituting the x and y values in the equations.
Plugging (2, 4) in equation 1, you get:
⇨ 4(2) – 6(4) = –16
8 – 24 = –16
–16 = –16 ✔
Plugging (2, 4) in equation 2, you get:
⇨ 8(2) + 2(4) = 24
16 + 8 = 24
24 = 24 ✔
Which Variable to Isolate When Solving a System with Substitution
It doesn’t matter which equation you choose or which variable you solve first; the solution of the system remains the same. Let’s take an example to illustrate this fact.
Solve this system by substitution.
–x + y = –4
4x – 3y = 10
We’ve shown you the step-by-step solution starting with x as well as y.
Isolate x First
Isolating x from equation 1, we have: ⇨ –x + y – y = –4 – y
[Subtracting y from both sides]
⇨ –x = –4 – y [Combining like terms]
⇨ x = 4 + y [Multiplying throughout by –1]
Now, substituting x in equation 2, we have:⇨ 4(4 + y) – 3y = 10
⇨ 4(4) + 4y – 3y = 10
[Applying distributive property]
⇨ 16 + 4y – 3y = 10
⇨ 16 + y = 10 [Combining like terms]
⇨ 16 + y – 16 = 10 – 16
[Subtracting 16 from both sides]
⇨ y = –6 [simplifying]
Plugging back y in x = 4 + y, we get:
⇨ x = 4 – 6
⇨ x = –2
Therefore, the solution set is (–2, –6).
Isolate y First
Isolating y from equation 1, we have:⇨ –x + y + x = –4 + x [Adding x to both sides]
⇨ y = –4 + x [Combining like terms]
Now, substituting y in equation 2, we have:⇨ 4x – 3(–4 + x) = 10
⇨ 4x – 3(–4) – 3x = 10
[Applying distributive property]
⇨ 4x + 12 – 3x = 10
⇨ x + 12 = 10 [Combining like terms]
⇨ x + 12 – 12 = 10 – 12
[Subtracting 12 from both sides]
⇨ x = –2 [simplifying]
Plugging back x in y = –4 + x, we get:
⇨ y = –4 – 2
⇨ y = –6
Therefore, the solution set is (–2, –6).
As you can see, the solution in both the cases is the same. So, just get started with a step that’s most convenient and simple.
The Gist of What We Learned So Far!
Solving a system of equations in two variables using substitution is easy and quick!
To solve by substitution, you need to follow a three-step process that involves isolation of a variable, substitution of the expression, and re-substitution of the value.
No matter which equation you choose and which variable you solve first; the solution of the system of equations stays the same.
Refine your practice with our free printable Solving Systems of Linear Equations worksheets!
