Transcript Elimination using Multiplication

Elimination using
Multiplication
Honors Math – Grade 8
Elimination using Multiplication

Some systems of
equations cannot be
solved simply by adding
or subtracting the
equations. In such cases,
one or both equations
must first be multiplied by
a number before the
system can be solved by
elimination.
 2x  y  6

3x  4 y  2
 2x  y  6

3x  4 y  2
1
Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply
one or both equations to change the coefficients to enable elimination.
Since the first equation has “–y” and the second equation has “+4y,” multiply the first equation
by 4.
1. Write the equations in column form and multiply.
4(2 x 
y6)
3x  4 y  2
Distribute
on both
sides of =
8 x  4 y  24
+3x  4 y  2
The y variable is
eliminated
because -4 + 4 = 0
11x  22
Solve the equation
2. Since the coefficients of y
are 4 and -4 (opposites), add
the equations together.
x=2
3. Now substitute x = 2 in either equation and solve.
The solution is (2, -2)
3x + 4y =
-2
3(2) + 4y =
-2
6 + 4y =
-2
4y =
-8
y =
-2
6 x  2 y  10

3x  7 y  19
2
Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply
one or both equations to change the coefficients to enable elimination.
Since the first equation has “6x” and the second equation has “3x,” multiply the second
equation by 2.
1. Write the equations in column form and multiply.
6 x  2 y  10
2(3 x  7 y
 19)
Distribute
on both
sides of =
6 x  2 y  10
- 6 x  14 y  38
2. Since the coefficients of x
are 6 and 6 (the same),
subtract the equations.
12 y  48
y=4
3. Now substitute y = 4 in either equation and solve.
The solution is (3, 4)
The x variable is
eliminated
because 6 - 6 = 0
Solve the equation
6x - 2y =
10
6x – 2(4) =
10
6x – 8 =
10
6x =
18
x =
3
3
Method 1
Eliminate x
3x  4 y  25

2 x  3 y  6
Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply
one or both equations to change the coefficients to enable elimination.
To eliminate x, think what is the LCM of 2 and 3?
1. Write the equations in column form and multiply.
2(3 x  4 y
 25)
3(2 x  3 y
 6)
2. Since the
coefficients of x are 6
and 6 (the same),
subtract the
equations.
6 x  8 y  50
- 6 x  9 y  18
Distribute
on both
sides of =
17 y  68
y = -4
3. Now substitute y = -4 in either equation and solve.
The solution is (-3, -4)
6
The x variable is
eliminated
because 6 - 6 = 0
Solve the equation
3x + 4y =
-25
3x + 4(-4) =
-25
3x – 16 =
-25
3x =
-9
x =
-3
3
Method 2
Eliminate y
3x  4 y  25

2 x  3 y  6
Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply
one or both equations to change the coefficients to enable elimination.
To eliminate y, think what is the LCM of 4 and 3?
1. Write the equations in column form and multiply.
3(3 x  4 y
 25)
4(2 x  3 y
 6)
2. Since the
coefficients of y are
12 and -12
(opposites), add the
equations.
9 x  12 y  75
+8 x  12 y  24
Distribute
on both
sides of =
17 x  51
x = -3
3. Now substitute x = -3 in either equation and solve.
The solution is (-3, -4)
12
The y variable is
eliminated because
12 + -12 = 0
Solve the equation
3x + 4y =
-25
3(-3) + 4y =
-25
-9 + 4y =
-25
4y =
-16
y =
-4
4
Method 1
Eliminate a
6a  2b  2

 4a  3b  8
Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply
one or both equations to change the coefficients to enable elimination.
To eliminate a, think what is the LCM of 6 and 4?
1. Write the equations in column form and multiply.
2( 6a  2b  2)
3( 4a  3b  8 )
2. Since the
coefficients of a are
12 and 12 (the same),
subtract the
equations.
Distribute
on both
sides of =
12a  4b  4
12a  9b  24
 5b  20
b=4
3. Now substitute b = 4 in either equation and solve.
The solution is (-1, 4)
12
The a variable is
eliminated because
12 - 12 = 0
Solve the equation
6a + 2b =
2
6a + 2(4) =
2
6a + 8 =
2
6a =
-6
a =
-1
4
Method 2
Eliminate b
6a  2b  2

 4a  3b  8
Adding or Subtracting the equations will not eliminate a variable. Therefore, we must multiply
one or both equations to change the coefficients to enable elimination.
To eliminate b, think what is the LCM of 2 and 3?
1. Write the equations in column form and multiply.
18a  6b  6
3( 6a  2b  2)
2( 4a  3b  8 )
2. Since the
coefficients of b are 6
and 6 (the same),
subtract the
equations.
Distribute
on both
sides of =
8a  6b  16
10a  10
a = -1
3. Now substitute a = -1 in either equation and solve.
The solution is (-1, 4)
6
The b variable is
eliminated because
6–6 =0
Solve the equation
6a + 2b =
2
6(-1) + 2b =
2
-6 + 2b =
2
2b =
8
b =
4