Home / United States / Math Classes / 8th Grade Math / Solving System of Linear Equations by Substitution Method
A system of linear equations is a collection of two or more linear equations that have multiple variables. Here we will learn how to solve a system of linear equations using the substitution method. We will also look at some solved examples to understand the steps involved in the substitution method....Read MoreRead Less
Equations are mathematical statements that have two algebraic expressions on both sides of the equal sign (=). It depicts the equality relationship between the expressions written on the left and the expressions written on the right. L.H.S = R.H.S (left hand side = right hand side) appears in every math equation. To find the value of an unknown variable representing an unknown quantity, equations can be solved.
Equations | Is it an equation? | |
---|---|---|
1. | y = 3x + 5 | yes |
2. | 7 + 2 = 10 - 1 | yes |
3. | \(y+x^2+4\) | no |
Linear equations are mathematical equations with 1 as the degree. The highest exponent of terms in such equations is 1. These can be further divided into one-variable linear equations, two-variable linear equations, three-variable linear equations, and so on. A linear equation with variables X and Y has the standard form aX + bY – c = 0, where a and b are the coefficients of X and Y, and c is the constant.
To solve a pair of linear equations in two variables, we need a minimum of two equations. Similarly, to solve three linear equations, we need a minimum of three equations.
The solution of simultaneous linear equations can be classified into two types: graphical and algebraic methods. The algebraic method includes the substitution method as one of its categories. In this article, you will learn about the substitution method and how to apply it to solve a linear equation through examples.
The algebraic method for solving simultaneous linear equations is the substitution method. The value of one variable from one equation is substituted in the other equation, as the name implies. A pair of linear equations is thus transformed into a single linear equation with only one variable, which can then be solved quickly.
The following steps can be used to solve a system of two equations with two unknown values. To solve the linear equation, a list of steps is provided below:
1. Expand the parenthesis to simplify the given equation.
2. For x or y, solve one of the equations.
3. In the other equation, substitute the step 2 solution.
4. Now, using basic arithmetic operations, solve the new equation.
5. Finally, solve the equation to determine the second variable’s value.
Example 1:
Solve the system by substitution.
y = 2x – 3 Equation 1
7x – 3y = 8 Equation 2
Solution:
Step 1: Notice that Equation 1 is solved for y. So, you can substitute 2x – 3 for y in Equation 2 to obtain an equation in one variable, x.
Then solve the equation to find the value of x.
7x – 3y = 8 Equation 2
7x – 32x – 3 = 8 Substitute 2x – 3 for y.
7x – 6x + 9 = 8 Distributive Property.
x + 9 = 8 Combining like terms.
x = -1 Subtract 9 from each side.
Step 2: Substitute -1 for x in Equation 1 and solve for y.
y = 2x – 3 Equation 1.
= 2 – 1 – 3 Substitute -1 for x.
= – 2 – 3 Multiply.
= – 5 Subtract.
The solution is (-1, -5).
Example 2:
Solve the system by substitution.
y = 2x – 5 Equation 1
8x – 5y = 13 Equation 2
Solution:
Step 1: Notice that Equation 1 is solved for y. So, you can substitute 2x + 5 for y in Equation 2 to obtain an equation in one variable, x.
Then solve the equation to find the value of x.
8x – 5y = 13 Equation 2
8x – 52x – 5 = 13 Substitute 2x – 5 for y.
8x – 10x + 25 = 13 Distributive property
-2x + 25 = 13 Combine like terms
-2x = -12 Subtract 25 from each side
x = 6 Divide each side by -2
Step 2: Substitute 6 for x in Equation 1 and solve for y.
y = 2x – 5 Equation 1
= 26 – 5 Substitute 6 for x .
= 12 – 5 Multiply.
= 7 Subtract.
The solution is (6, 7).
Example 3:
You’re throwing a birthday bash. You spend $130.00 on a total of 60 turkey and veggie burgers. The turkey burger costs $3.00 and the veggie burger costs $2.00. How many of each burger do you plan to purchase?
Big sandwich – hamburger burger with beef, red onion, tomato and fried bacon.
Solution:
To write a system of linear equations, use a verbal model. Let x stand for the number of turkey burgers, and y for the number of veggie burgers.
Number of turkey burgers, x + Number of veggie burgers, y = Total number of burgers
Cost per turkey burger Number of turkey burgers, x + Cost per veggie burger Number of veggie burgers, y ) = Total cost
The system is x + y = 60 Equation 1
3x + 2y = 130 Equation 2
Step 1: One solution method is to rewrite Equation 1 as x = 60 – y. Then substitute 60 – y for x in Equation 2 and solve to find the value of y.
3x + 2y = 130 Equation 2
3(60-y) + 2y = 130 Substitute 60-y for x.
180 – 3y + 2y = 130 Distributive Property.
180 – y = 130 Combine like terms
-y = -50 . Subtract 20 from each side.
y = 50 Divide each side by -1.
Step 2: Substitute 50 for y in Equation 1 and solve for x.
x + y = 60 Equation 1.
x + 50 = 60 Substitute 50 for y.
x = 10 Subtract 50 from each side.
You buy 10 turkey burgers and 50 veggie burgers.
The substitution method is commonly used in mathematics to solve a system of equations. Solve the equation for one variable first, then substitute the value of the variable in the other equation.
The substitution method has the advantage of giving exact values for the variables (x and y) that correspond to the point of intersection.
The substitution method includes the following steps:
Step-1: Solve one of the equations for the variable x or for the variable y.
Step-2: Now, in the other equation, substitute the solution from step one.
Step-3: Finally, solve the equation to determine the other variable’s value.