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Transcript 
Solving systems of equations
with 2 variables
Word problems
(Number Problems)
1) The sum of two numbers is 72. Their
difference is 40. Find the numbers.
The sum of two numbers is 72.
x + y = 72
Their difference is 40.
x – y = 40
1) The sum of two numbers is 72. Their
difference is 40. Find the numbers.
x + y = 72
x – y = 40
Which method should be used to solve
this system of equations?
a) Substitution Method
b) Elimination (Addition) Method
1) The sum of two numbers is 72. Their
difference is 40. Find the numbers.
Solve using the Elimination (Addition) Method
x + y = 72
x – y = 40
2x
= 112
x
= 56
Back substitute
56 + y = 72
The numbers
are 56 and 16.
56 + y + (-56) = 72 + (-56)
y = 16
2) The sum of two numbers is 21. Their
difference is 13. Find the numbers.
Solve using the Elimination (Addition) Method
x + y = 21
x – y = 13
2x
= 34
x
= 17
Back substitute
17 + y = 21
The numbers
are 17 and 4.
17 + y + (-17) = 21 + (-17)
y =4
3) The sum of two numbers is 27. One number
is three more than the other. Find the numbers.
The sum of two numbers is 27.
x + y = 27
One number is 3 more than the other.
y = x+3
3) The sum of two numbers is 27. One number
is three more than the other. Find the numbers.
x + y = 27
y = x+3
Which method should be used to solve
this system of equations?
a) Substitution Method
b) Elimination (Addition) Method
3) The sum of two numbers is 27. One number
is three more than the other. Find the numbers.
Solve using the Substitution Method
x + y = 27
y = x+3
x + (x + 3) = 27
2x + 3 = 27
2x + 3 + (-3) = 27 + (-3)
2x = 24
x = 12
The numbers
are 12 and 15.
Back substitution
y=x+3
y = 12 + 3
y = 15
4) A 36-ft rope is cut into two pieces. One
piece is three times the other. Find the length
of each piece.
Solve using the Substitution Method
x + y = 36
y = 3x
x + (3x) = 36
4x = 36
x=9
The pieces are
9 ft and 27 ft.
Back substitution
y = 3x
y = 3(9)
y = 27
5) The difference between two numbers is 3.
The larger number is one more than twice the
smaller number. Find the numbers.
Solve using the Substitution Method
L–S=3
L = 2S + 1
The numbers
are 2 and 5.
(2S + 1) – S = 3
Back substitution
S+1 =3
L = 2S + 1
L = 2(2) + 1
S + 1 + (-1) = 3 + (-1)
L=5
S=2
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