Two Variable Systems of Equations
Joseph Lee
Metropolitan Community College
Joseph Lee
Two Variable Systems of Equations
Example 1.
Solve the following system of equations by graphing.
3x + y = 1
−x + y = −3
y
x
Solution:
Joseph Lee
Two Variable Systems of Equations
Example 1.
Solve the following system of equations by graphing.
3x + y = 1
−x + y = −3
y
x
(1, −2)
Solution:
Joseph Lee
Two Variable Systems of Equations
Example 1.
Solve the following system of equations by graphing.
3x + y = 1
−x + y = −3
y
x
(1, −2)
Solution: (1, −2)
Joseph Lee
Two Variable Systems of Equations
Example 2.
Solve the following system of equations by graphing.
2x + 3y = 12
y = − 23 x + 4
y
x
Solution:
Joseph Lee
Two Variable Systems of Equations
Example 2.
Solve the following system of equations by graphing.
2x + 3y = 12
y = − 23 x + 4
y
x
Solution:
Joseph Lee
Two Variable Systems of Equations
Example 2.
Solve the following system of equations by graphing.
2x + 3y = 12
y = − 23 x + 4
y
x
Solution: All points on the line 2x + 3y = 12
Joseph Lee
Two Variable Systems of Equations
Example 3.
Solve the following system of equations by graphing.
y = −3x + 4
3x + y = 1
y
x
Solution:
Joseph Lee
Two Variable Systems of Equations
Example 3.
Solve the following system of equations by graphing.
y = −3x + 4
3x + y = 1
y
x
Solution:
Joseph Lee
Two Variable Systems of Equations
Example 3.
Solve the following system of equations by graphing.
y = −3x + 4
3x + y = 1
y
x
Solution: No Solution
Joseph Lee
Two Variable Systems of Equations
Definition: Dependent, Independent, Inconsistent Systems
A system of equations is called inconsistent if it has no solutions.
A system of equations is called dependent if it has infinitely many
solutions.
A system of equations is called independent if it has a single
solution.
Joseph Lee
Two Variable Systems of Equations
Example 4.
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Joseph Lee
Two Variable Systems of Equations
Example 4.
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our first
equation x − 2y = 8 and solve for x.
Joseph Lee
Two Variable Systems of Equations
Example 4.
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our first
equation x − 2y = 8 and solve for x.
x − 2y = 8
Joseph Lee
Two Variable Systems of Equations
Example 4.
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our first
equation x − 2y = 8 and solve for x.
x − 2y = 8
x = 2y + 8
Joseph Lee
Two Variable Systems of Equations
Example 4.
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our first
equation x − 2y = 8 and solve for x.
x − 2y = 8
x = 2y + 8
Now, we may substitute x = 2y + 8 into our other equation
4x + y = 5.
Joseph Lee
Two Variable Systems of Equations
Example 4.
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our first
equation x − 2y = 8 and solve for x.
x − 2y = 8
x = 2y + 8
Now, we may substitute x = 2y + 8 into our other equation
4x + y = 5.
4x + y = 5
Joseph Lee
Two Variable Systems of Equations
Example 4.
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our first
equation x − 2y = 8 and solve for x.
x − 2y = 8
x = 2y + 8
Now, we may substitute x = 2y + 8 into our other equation
4x + y = 5.
4x + y = 5
4(2y + 8) + y = 5
Joseph Lee
Two Variable Systems of Equations
Example 4.
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our first
equation x − 2y = 8 and solve for x.
x − 2y = 8
x = 2y + 8
Now, we may substitute x = 2y + 8 into our other equation
4x + y = 5.
4x + y = 5
4(2y + 8) + y = 5
8y + 32 + y = 5
Joseph Lee
Two Variable Systems of Equations
Example 4.
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our first
equation x − 2y = 8 and solve for x.
x − 2y = 8
x = 2y + 8
Now, we may substitute x = 2y + 8 into our other equation
4x + y = 5.
4x + y = 5
4(2y + 8) + y = 5
8y + 32 + y = 5
9y = −27
Joseph Lee
Two Variable Systems of Equations
Example 4.
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our first
equation x − 2y = 8 and solve for x.
x − 2y = 8
x = 2y + 8
Now, we may substitute x = 2y + 8 into our other equation
4x + y = 5.
4x + y = 5
4(2y + 8) + y = 5
8y + 32 + y = 5
9y = −27
y = −3
Joseph Lee
Two Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. (Continued)
4x + y
4(2y + 8) + y
8y + 32 + y
9y
y
Joseph Lee
=5
=5
=5
= −27
= −3
Two Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. (Continued)
4x + y
4(2y + 8) + y
8y + 32 + y
9y
y
=5
=5
=5
= −27
= −3
Since y = −3, we may substitute that value back into any
equation to solve for x.
Joseph Lee
Two Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. (Continued)
4x + y
4(2y + 8) + y
8y + 32 + y
9y
y
=5
=5
=5
= −27
= −3
Since y = −3, we may substitute that value back into any
equation to solve for x.
x = 2y + 8
Joseph Lee
Two Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. (Continued)
4x + y
4(2y + 8) + y
8y + 32 + y
9y
y
=5
=5
=5
= −27
= −3
Since y = −3, we may substitute that value back into any
equation to solve for x.
x = 2y + 8
x = 2(−3) + 8
Joseph Lee
Two Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. (Continued)
4x + y
4(2y + 8) + y
8y + 32 + y
9y
y
=5
=5
=5
= −27
= −3
Since y = −3, we may substitute that value back into any
equation to solve for x.
x = 2y + 8
x = 2(−3) + 8
x =2
Joseph Lee
Two Variable Systems of Equations
Example 4. (Continued)
Solve the following system of equations using substitution.
x − 2y = 8
4x + y = 5
Solution. (Continued)
4x + y
4(2y + 8) + y
8y + 32 + y
9y
y
=5
=5
=5
= −27
= −3
Since y = −3, we may substitute that value back into any
equation to solve for x.
x = 2y + 8
x = 2(−3) + 8
x =2
Thus, (2, −3) is the solution of this system of equations.
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our second
equation 4x + 2y = 2 and solve for y .
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our second
equation 4x + 2y = 2 and solve for y .
4x + 2y = 2
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our second
equation 4x + 2y = 2 and solve for y .
4x + 2y = 2
2y = −4x + 2
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our second
equation 4x + 2y = 2 and solve for y .
4x + 2y = 2
2y = −4x + 2
y = −2x + 1
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our second
equation 4x + 2y = 2 and solve for y .
4x + 2y = 2
2y = −4x + 2
y = −2x + 1
Now, we may substitute y = −2x + 1 into our other equation
5x + 3y = 6.
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our second
equation 4x + 2y = 2 and solve for y .
4x + 2y = 2
2y = −4x + 2
y = −2x + 1
Now, we may substitute y = −2x + 1 into our other equation
5x + 3y = 6.
5x + 3y = 6
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our second
equation 4x + 2y = 2 and solve for y .
4x + 2y = 2
2y = −4x + 2
y = −2x + 1
Now, we may substitute y = −2x + 1 into our other equation
5x + 3y = 6.
5x + 3y = 6
5x + 3(−2x + 1) = 6
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our second
equation 4x + 2y = 2 and solve for y .
4x + 2y = 2
2y = −4x + 2
y = −2x + 1
Now, we may substitute y = −2x + 1 into our other equation
5x + 3y = 6.
5x + 3y = 6
5x + 3(−2x + 1) = 6
5x − 6x + 3 = 6
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our second
equation 4x + 2y = 2 and solve for y .
4x + 2y = 2
2y = −4x + 2
y = −2x + 1
Now, we may substitute y = −2x + 1 into our other equation
5x + 3y = 6.
5x + 3y = 6
5x + 3(−2x + 1) = 6
5x − 6x + 3 = 6
−x =3
Joseph Lee
Two Variable Systems of Equations
Example 5.
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by substitution, we will take one of our
equations and solve for one of our variables. Let’s take our second
equation 4x + 2y = 2 and solve for y .
4x + 2y = 2
2y = −4x + 2
y = −2x + 1
Now, we may substitute y = −2x + 1 into our other equation
5x + 3y = 6.
5x + 3y = 6
5x + 3(−2x + 1) = 6
5x − 6x + 3 = 6
−x =3
x = −3
Joseph Lee
Two Variable Systems of Equations
Example 5. (Continued)
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
5x + 3y
5x + 3(−2x + 1)
5x − 6x + 3
−x
x
Joseph Lee
=6
=6
=6
=3
= −3
Two Variable Systems of Equations
Example 5. (Continued)
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
5x + 3y
5x + 3(−2x + 1)
5x − 6x + 3
−x
x
=6
=6
=6
=3
= −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
Joseph Lee
Two Variable Systems of Equations
Example 5. (Continued)
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
5x + 3y
5x + 3(−2x + 1)
5x − 6x + 3
−x
x
=6
=6
=6
=3
= −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
y = −2x + 1
Joseph Lee
Two Variable Systems of Equations
Example 5. (Continued)
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
5x + 3y
5x + 3(−2x + 1)
5x − 6x + 3
−x
x
=6
=6
=6
=3
= −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
y = −2x + 1
y = −2(−3) + 1
Joseph Lee
Two Variable Systems of Equations
Example 5. (Continued)
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
5x + 3y
5x + 3(−2x + 1)
5x − 6x + 3
−x
x
=6
=6
=6
=3
= −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
y = −2x + 1
y = −2(−3) + 1
y =7
Joseph Lee
Two Variable Systems of Equations
Example 5. (Continued)
Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
5x + 3y
5x + 3(−2x + 1)
5x − 6x + 3
−x
x
=6
=6
=6
=3
= −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
y = −2x + 1
y = −2(−3) + 1
y =7
Thus, (−3, 7) is the solution of this system of equations.
Joseph Lee
Two Variable Systems of Equations
Example 6.
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Joseph Lee
Two Variable Systems of Equations
Example 6.
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
Joseph Lee
Two Variable Systems of Equations
Example 6.
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
x − 2y = 8
4x + y = 5
Joseph Lee
Two Variable Systems of Equations
Example 6.
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
x − 2y = 8
4x + y = 5
x − 2y = 8
2(4x + y ) = (5)2
Joseph Lee
Two Variable Systems of Equations
Example 6.
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
x − 2y = 8
4x + y = 5
x − 2y = 8
2(4x + y ) = (5)2
x − 2y = 8
8x + 2y = 10
Joseph Lee
Two Variable Systems of Equations
Example 6.
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
x − 2y = 8
4x + y = 5
x − 2y = 8
2(4x + y ) = (5)2
x − 2y = 8
8x + 2y = 10
9x = 18
Joseph Lee
Two Variable Systems of Equations
Example 6.
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
x − 2y = 8
4x + y = 5
x − 2y = 8
2(4x + y ) = (5)2
x − 2y = 8
8x + 2y = 10
9x = 18
x =2
Joseph Lee
Two Variable Systems of Equations
Example 6. (Continued)
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. (Continued)
x =2
Joseph Lee
Two Variable Systems of Equations
Example 6. (Continued)
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. (Continued)
x =2
Since x = 2, we may substitute that value back into any equation
to solve for y .
Joseph Lee
Two Variable Systems of Equations
Example 6. (Continued)
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. (Continued)
x =2
Since x = 2, we may substitute that value back into any equation
to solve for y .
4x + y = 5
Joseph Lee
Two Variable Systems of Equations
Example 6. (Continued)
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. (Continued)
x =2
Since x = 2, we may substitute that value back into any equation
to solve for y .
4x + y = 5
4(2) + y = 5
Joseph Lee
Two Variable Systems of Equations
Example 6. (Continued)
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. (Continued)
x =2
Since x = 2, we may substitute that value back into any equation
to solve for y .
4x + y = 5
4(2) + y = 5
8+y =5
Joseph Lee
Two Variable Systems of Equations
Example 6. (Continued)
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. (Continued)
x =2
Since x = 2, we may substitute that value back into any equation
to solve for y .
4x + y = 5
4(2) + y = 5
8+y =5
y = −3
Joseph Lee
Two Variable Systems of Equations
Example 6. (Continued)
Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
Solution. (Continued)
x =2
Since x = 2, we may substitute that value back into any equation
to solve for y .
4x + y = 5
4(2) + y = 5
8+y =5
y = −3
Thus, (2, −3) is the solution of this system of equations.
Joseph Lee
Two Variable Systems of Equations
Example 7.
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Joseph Lee
Two Variable Systems of Equations
Example 7.
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
Joseph Lee
Two Variable Systems of Equations
Example 7.
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
5x + 3y = 6
4x + 2y = 2
Joseph Lee
Two Variable Systems of Equations
Example 7.
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
5x + 3y = 6
4x + 2y = 2
2(5x + 3y ) = (6)2
−3(4x + 2y ) = (2)(−3)
Joseph Lee
Two Variable Systems of Equations
Example 7.
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
5x + 3y = 6
4x + 2y = 2
2(5x + 3y ) = (6)2
−3(4x + 2y ) = (2)(−3)
10x + 6y = 12
−12x − 6y = −6
Joseph Lee
Two Variable Systems of Equations
Example 7.
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
5x + 3y = 6
4x + 2y = 2
2(5x + 3y ) = (6)2
−3(4x + 2y ) = (2)(−3)
10x + 6y = 12
−12x − 6y = −6
−2x = 6
Joseph Lee
Two Variable Systems of Equations
Example 7.
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by elimination, we will eliminate one of our
variables. Let’s eliminate y .
5x + 3y = 6
4x + 2y = 2
2(5x + 3y ) = (6)2
−3(4x + 2y ) = (2)(−3)
10x + 6y = 12
−12x − 6y = −6
−2x = 6
x = −3
Joseph Lee
Two Variable Systems of Equations
Example 7. (Continued)
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
x = −3
Joseph Lee
Two Variable Systems of Equations
Example 7. (Continued)
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
x = −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
Joseph Lee
Two Variable Systems of Equations
Example 7. (Continued)
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
x = −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
4x + 2y = 2
Joseph Lee
Two Variable Systems of Equations
Example 7. (Continued)
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
x = −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
4x + 2y = 2
4(−3) + 2y = 2
Joseph Lee
Two Variable Systems of Equations
Example 7. (Continued)
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
x = −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
4x + 2y = 2
4(−3) + 2y = 2
− 12 + 2y = 2
Joseph Lee
Two Variable Systems of Equations
Example 7. (Continued)
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
x = −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
4x + 2y
4(−3) + 2y
− 12 + 2y
2y
Joseph Lee
=2
=2
=2
= 14
Two Variable Systems of Equations
Example 7. (Continued)
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
x = −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
4x + 2y
4(−3) + 2y
− 12 + 2y
2y
y
Joseph Lee
=2
=2
=2
= 14
=7
Two Variable Systems of Equations
Example 7. (Continued)
Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. (Continued)
x = −3
Since x = −3, we may substitute that value back into any
equation to solve for y .
4x + 2y
4(−3) + 2y
− 12 + 2y
2y
y
=2
=2
=2
= 14
=7
Thus, (−3, 7) is the solution of this system of equations.
Joseph Lee
Two Variable Systems of Equations
Example 8.
Solve the following system of equations using elimination.
4x + 6y = 24
y = − 32 x + 4
Joseph Lee
Two Variable Systems of Equations
Example 8.
Solve the following system of equations using elimination.
4x + 6y = 24
y = − 32 x + 4
Solution. First, let’s get rid of the fraction. Then we will move
our variables over to the left side. Then we may choose a variable
to eliminate.
Joseph Lee
Two Variable Systems of Equations
Example 8.
Solve the following system of equations using elimination.
4x + 6y = 24
y = − 32 x + 4
Solution. First, let’s get rid of the fraction. Then we will move
our variables over to the left side. Then we may choose a variable
to eliminate.
4x + 6y = 24
3(y ) = − 23 x + 4 3
Joseph Lee
Two Variable Systems of Equations
Example 8.
Solve the following system of equations using elimination.
4x + 6y = 24
y = − 32 x + 4
Solution. First, let’s get rid of the fraction. Then we will move
our variables over to the left side. Then we may choose a variable
to eliminate.
4x + 6y = 24
3(y ) = − 23 x + 4 3
4x + 6y = 24
3y = −2x + 12
Joseph Lee
Two Variable Systems of Equations
Example 8.
Solve the following system of equations using elimination.
4x + 6y = 24
y = − 32 x + 4
Solution. First, let’s get rid of the fraction. Then we will move
our variables over to the left side. Then we may choose a variable
to eliminate.
4x + 6y = 24
3(y ) = − 23 x + 4 3
4x + 6y = 24
3y = −2x + 12
4x + 6y = 24
2x + 3y = 12
Joseph Lee
Two Variable Systems of Equations
Example 8. (Continued)
Solution. (Continued)
4x + 6y = 24
2x + 3y = 12
Joseph Lee
Two Variable Systems of Equations
Example 8. (Continued)
Solution. (Continued)
4x + 6y = 24
2x + 3y = 12
4x + 6y = 24
−2(2x + 3y ) = (12)(−2)
Joseph Lee
Two Variable Systems of Equations
Example 8. (Continued)
Solution. (Continued)
4x + 6y = 24
2x + 3y = 12
4x + 6y = 24
−2(2x + 3y ) = (12)(−2)
4x + 6y = 24
−4x − 6y = −24
Joseph Lee
Two Variable Systems of Equations
Example 8. (Continued)
Solution. (Continued)
4x + 6y = 24
2x + 3y = 12
4x + 6y = 24
−2(2x + 3y ) = (12)(−2)
4x + 6y = 24
−4x − 6y = −24
0 =0
Joseph Lee
Two Variable Systems of Equations
Example 8. (Continued)
Solution. (Continued)
4x + 6y = 24
2x + 3y = 12
4x + 6y = 24
−2(2x + 3y ) = (12)(−2)
4x + 6y = 24
−4x − 6y = −24
0 =0
Arriving at this identity, we may conclude that this system is
dependent. There are infinitely many solutions, or more
specifically, the solution is
Joseph Lee
Two Variable Systems of Equations
Example 8. (Continued)
Solution. (Continued)
4x + 6y = 24
2x + 3y = 12
4x + 6y = 24
−2(2x + 3y ) = (12)(−2)
4x + 6y = 24
−4x − 6y = −24
0 =0
Arriving at this identity, we may conclude that this system is
dependent. There are infinitely many solutions, or more
specifically, the solution is any point on the line 4x + 6y = 24.
Joseph Lee
Two Variable Systems of Equations
Example 9.
Solve the following system of equations using elimination.
y = −3x + 4
6x + 2y = −2
Joseph Lee
Two Variable Systems of Equations
Example 9.
Solve the following system of equations using elimination.
y = −3x + 4
6x + 2y = −2
Solution. First, we will move our variables over to the left side.
Then we may choose a variable to eliminate.
Joseph Lee
Two Variable Systems of Equations
Example 9.
Solve the following system of equations using elimination.
y = −3x + 4
6x + 2y = −2
Solution. First, we will move our variables over to the left side.
Then we may choose a variable to eliminate.
y = −3x + 4
6x + 2y = −2
Joseph Lee
Two Variable Systems of Equations
Example 9.
Solve the following system of equations using elimination.
y = −3x + 4
6x + 2y = −2
Solution. First, we will move our variables over to the left side.
Then we may choose a variable to eliminate.
y = −3x + 4
6x + 2y = −2
3x + y = 4
6x + 2y = −2
Joseph Lee
Two Variable Systems of Equations
Example 9.
Solve the following system of equations using elimination.
y = −3x + 4
6x + 2y = −2
Solution. First, we will move our variables over to the left side.
Then we may choose a variable to eliminate.
y = −3x + 4
6x + 2y = −2
3x + y = 4
6x + 2y = −2
−2(3x + y ) = (4)(−2)
6x + 2y = −2
Joseph Lee
Two Variable Systems of Equations
Example 9. (Continued)
Solution. (Continued)
3x + y = 4
6x + 2y = −2
Joseph Lee
Two Variable Systems of Equations
Example 9. (Continued)
Solution. (Continued)
3x + y = 4
6x + 2y = −2
−2(3x + y ) = (4)(−2)
6x + 2y = −2
Joseph Lee
Two Variable Systems of Equations
Example 9. (Continued)
Solution. (Continued)
3x + y = 4
6x + 2y = −2
−2(3x + y ) = (4)(−2)
6x + 2y = −2
−6x − 2y = −8
6x + 2y = −2
Joseph Lee
Two Variable Systems of Equations
Example 9. (Continued)
Solution. (Continued)
3x + y = 4
6x + 2y = −2
−2(3x + y ) = (4)(−2)
6x + 2y = −2
−6x − 2y = −8
6x + 2y = −2
0 = −10
Joseph Lee
Two Variable Systems of Equations
Example 9. (Continued)
Solution. (Continued)
3x + y = 4
6x + 2y = −2
−2(3x + y ) = (4)(−2)
6x + 2y = −2
−6x − 2y = −8
6x + 2y = −2
0 = −10
Arriving at this contradiction, we may conclude that this system is
inconsistent. There is
Joseph Lee
Two Variable Systems of Equations
Example 9. (Continued)
Solution. (Continued)
3x + y = 4
6x + 2y = −2
−2(3x + y ) = (4)(−2)
6x + 2y = −2
−6x − 2y = −8
6x + 2y = −2
0 = −10
Arriving at this contradiction, we may conclude that this system is
inconsistent. There is no solution.
Joseph Lee
Two Variable Systems of Equations