Substitution in algebra 1
Introduction to substitution in algebra 1:
System of simultaneous equations can be solved by using substitution method. Steps used to solve algebra 1 substitution method:
Step 1: Let us take any one of the equation from the given two equations. From this equation, write one variable in terms of the other variable.
Step 2: Next, substitute this in another one equation to get the single variable equation. Simplify and then find a one variable value.
Step 3: Substitute this variable value in any one of the equation, so we get the other another one variable value.
Now, we are going to see some of the problems in substitution in algebra 1.
Problems on substitution in algebra 1:
Example problem 1:
Solve the following two linear equation by substitution method:
5x + y = 3.5---------Equation (1)
x – 2y = 4---------Equation (2)
Solution:
Step 1: Let us consider the equation (1)
5x + y = 3.5
Subtract 5x on both sides of the equation
5x + y - 5x = 3.5 – 5x
y=-5x + 3.5------------------------Equation (3)
Step 2: Substitute this value of y in Equation (2). We get
x – 2y = 4
x – 2(-5x + 3.5) = 4
x + 10x – 7 = 4
11x – 7 = 4
Add 7 on both sides of the equation
11x – 7 + 7 = 4 + 7
11x = 11
Divide by 11 on both sides of the equation
11x / 11 = 11 / 11
x = 1
Step 3: Plugging this value of x in Equation (3), we get
y = -5x + 3.5
y = -5 + 3.5
y = -1.5
So, the solution of the given two linear equations is (1, -1.5).
Additional problems on substitution in algebra 1:
Example problem 2:
Solve the following two linear equations by substitution method:
y = 1.5x ------------Equation (1)
x + y = 4------------Equation (2)
Solution:
Step 1: Let us consider the equation (1)
y = 1.5x
Step 2: Substitute the value of y in Equation (2). We get
x + y = 5
x + (1.5x) = 5
x + 1.5x = 5
2.5x = 5
Divide by 2.5 on both sides of the equation
2.5x / 2.5 = 5 / 2.5
x = 2
Step 3: Plugging this value of x in Equation (1), we get
y= 1.5x
y = 1.5(2)
y = 3
So, the solution of the given two linear equations is (2, 3).
System of simultaneous equations can be solved by using substitution method. Steps used to solve algebra 1 substitution method:
Step 1: Let us take any one of the equation from the given two equations. From this equation, write one variable in terms of the other variable.
Step 2: Next, substitute this in another one equation to get the single variable equation. Simplify and then find a one variable value.
Step 3: Substitute this variable value in any one of the equation, so we get the other another one variable value.
Now, we are going to see some of the problems in substitution in algebra 1.
Problems on substitution in algebra 1:
Example problem 1:
Solve the following two linear equation by substitution method:
5x + y = 3.5---------Equation (1)
x – 2y = 4---------Equation (2)
Solution:
Step 1: Let us consider the equation (1)
5x + y = 3.5
Subtract 5x on both sides of the equation
5x + y - 5x = 3.5 – 5x
y=-5x + 3.5------------------------Equation (3)
Step 2: Substitute this value of y in Equation (2). We get
x – 2y = 4
x – 2(-5x + 3.5) = 4
x + 10x – 7 = 4
11x – 7 = 4
Add 7 on both sides of the equation
11x – 7 + 7 = 4 + 7
11x = 11
Divide by 11 on both sides of the equation
11x / 11 = 11 / 11
x = 1
Step 3: Plugging this value of x in Equation (3), we get
y = -5x + 3.5
y = -5 + 3.5
y = -1.5
So, the solution of the given two linear equations is (1, -1.5).
Additional problems on substitution in algebra 1:
Example problem 2:
Solve the following two linear equations by substitution method:
y = 1.5x ------------Equation (1)
x + y = 4------------Equation (2)
Solution:
Step 1: Let us consider the equation (1)
y = 1.5x
Step 2: Substitute the value of y in Equation (2). We get
x + y = 5
x + (1.5x) = 5
x + 1.5x = 5
2.5x = 5
Divide by 2.5 on both sides of the equation
2.5x / 2.5 = 5 / 2.5
x = 2
Step 3: Plugging this value of x in Equation (1), we get
y= 1.5x
y = 1.5(2)
y = 3
So, the solution of the given two linear equations is (2, 3).