System of Equations
= a set of equations describing the same situation.
2x − 3y = 3
4x + y = 9
2x − 3y + 5z = 1
4x + y − z = 2
x − 2y + 3z = −5
If there are two variables you need two equations to be able to solve it.
If there are three variables you need three equations to be able to solve it.
2x − 3y = 6 → 1
x + 2y = 10 → 2
Solving By Graphing.
Graph each equation using intercepts.
Equation1
x-intercept = 3
y-intercept = −2
Equation 2
x-intercept = 10
y-intercept = 5
The intersection of the two graphs shows the values that fit both equations
(6, 2)
CHECK YOUR SOLUTION BY SUBSTITUTING AND BY USING THE
INTERSECTION FUNCTION ON YOUR CALCULATOR.
Solving By Graphing.
x = 12 y + 1 → 1
x − 2y = −2 → 2
Equation1
x-intercept = 1
y-intercept = −2
Equation 2
x-intercept = −2
y-intercept = 1
Solving By Substitution.
y − 2x = −5 → 1
3y − x = 5 → 2
Transpose one of the equations to give a term that you can substitute into the other
equation.
1 → y = 2x − 5
2 → 3(2x − 5) − x = 5
6x − 15 − x = 5
5x = 20
x=4
Substitute into one of the equations to find y.
1 → y − 2(4) = −5
Solution is (2, 2)
Solving By Substitution.
3x + 2y = 1 → 1
5x + 4y = 3 → 2
Transpose one of the equations to give a term that you can substitute into the other
equation.
1 → 2y = 1 − 3x
2 → 5x + 2(1 − 3x ) = 3
5x + 2 − 6x = 3
2−x =3
x = −1
Solving By Elimination.
If you’re subtracting
think about which
order makes it easier
2-1 or 1-2
1 → 3(−1) + 2y = 1
1 → −1 + n = −7
n = −6
Solution is m = −1; n = −6
→y=2
Solution is (−1, 2)
m + n = −7 → 1
3m + n = −9 → 2
2 − 1 → 2m
= −2
m = −1
Substitute into one of the first equations to find the other variable
NOTE: Always substitute into one of the original equations not one that you have
transposed.
CHECK YOUR SOLUTION.
Solution is (4,3)
CHECK YOUR SOLUTION.
CHECK YOUR SOLUTION.
→y=3
CHECK YOUR SOLUTION.
1
Sometimes
something has to
be done before
you can eliminate
one of the terms.
If you have
trouble with
subtracting
negative integers
it is easier to set it
up so that you
can eliminate by
adding rather
than subtracting.
m − n = 7 →1
4m − 5n = 25 → 2
1× (−5) → −5m + 5n = −35 → 3
2 + 3 → −m
= −10
m = 10
substitute into 1
10 − n = 7
n =3
Solution is m = 10; n = 3
CHECK YOUR SOLUTION.
5m − 2n = 0 → 1
2m − 3n = −11 → 2
1× 2 → 10m − 4n = 0 → 3
2 × 5 → 10m − 15n = −55 → 4
3− 4 →
11n = 55
n =5
substitute into 1
5m − 10 = 0
5m = 10
m=2
CHECK YOUR SOLUTION
USING THE ORIGINAL
EQUATIONS.
Solution is m = 2; n = 5
It took 3 hours to row a boat upstream to reach a café for lunch.
The return trip downstream took 2 hours. If the current speed was
3 km/hr find the speed of the boat in still water.
Let B = boat speed in still water, D = distance travelled
Against the current 3(B − 3) = D
With the current 2(B + 3) = D
3B − 9 = D → 1
2B + 6 = D→ 2
1− 2 → B − 15 = 0
B = 15
CHECK YOUR SOLUTION
USING THE ORIGINAL
WORDING NOT YOUR
EQUATIONS.
The speed of the boat is 15 km / hr
3x − 2y = 10 → 1
5x + 3y = 4 → 2
1× 3 → 9x − 6y = 30 → 3
2 × 2 → 10x + 6y = 8 → 4
3 + 4 → 19x
= 38
x=2
substitute into 2
10 + 3y = 4
3y = −6
y = −2
CHECK YOUR SOLUTION
USING THE ORIGINAL
EQUATIONS.
Solution is (2, −2)
Motion Questions:
An airplane took 2 hours to fly 600 km against a head wind. The
return trip with the wind took 12/3 hours. Find the speed of the
plane in still air and the wind speed.
Let P = plane speed in still air, W = speed of the wind
These questions are based on the relationship that time × speed = distance
Against the wind 2(P − W) = 600
P − W = 300 → 1
With the wind 1 (P + W) = 600
P + W = 360 → 2
2
3
1 + 2 → 2P = 660
P = 330
substitute into 1
330 − W = 300
W = 30
speed of the wind is 30 km / hr
The speed of the plane in still air is 330km/hr
CHECK YOUR SOLUTION
USING THE ORIGINAL
WORDING NOT YOUR
The
EQUATIONS.
Digit Questions:
The sum of the digits of a two-digit number is 14. If the
number represented by reversing the digits is subtracted from the original number,
the result is 18. Find the original number.
Let the original number be xy where
x = the first digit in the number, y = the second digit in the number
The numerical value of this number is 10 ⋅ x + y
The new number with the digits reversed will be yx
The numerical value of this number is 10 ⋅ y + x
The sum of the digits of a two-digit number is 14 → 1
The original number minus the new number is 18 → 2
x + y = 14 → 1
(10 ⋅ x + y) − (10 ⋅ y + x) = 18 → 2
10x + y − 10y − x = 18
9x − 9y = 18
x−y = 2→3
1 + 3 → 2x = 16
x= 8
CHECK YOUR SOLUTION
USING THE ORIGINAL
substitute into 1 → y = 6
WORDING NOT YOUR
EQUATIONS.
The original number is 86
2
The sum of the numerator and denominator of a fraction is 20. If five is subtracted
from both the numerator and the denominator the fraction can be simplified to 3/2
Find the original fraction.
Let the original fraction be
n + d = 20 → 1
n−5 3
=
d −5 2
n
d
2n − 10 = 3d − 15
2n − 3d = −5 → 2
1 → n = 20 − d
substitute into 2 → 40 − 2d − 3d = −5
40 − 5d = −5
−5d = −45
d=9
substitute into 1 → n = 11
CHECK YOUR SOLUTION
USING THE ORIGINAL
WORDING NOT YOUR
EQUATIONS.
The original fraction was = 11
9
How much 42% acid solution needs to be added to a 56% acid solution to make 21L of
48% acid solution.
0.42x + 0.56y = 0.48 × 21
42x + 56y = 1008 ÷ both sides by14
3x + 4y = 72 → 1
x + y = 21 → 2
2 → x = 21 − y
substitute into 1
3(21 − y) + 4y = 72
63 − 3y + 4y = 72
y=9
sub into 2 → x = 12
Mixture Questions: There were 200 tickets sold for a basketball game. If adult
tickets $9 and child tickets are $4.50. If the total money collected was $1485 find the
number of each type of ticket sold.
Let A = the number of adult tickets sold
Let C = the number of child tickets sold.
There were 200 tickets sold → 1
Adult tickets are $9 and child tickets are $4.50.
The total money collected was $1485 → 2
using number of tickets
A + C = 200 → 1
using money collected 9A + 4.5C = 1485
2A + C = 330 → 2
2 − 1 → A = 130
substitute into 1 → C = 70
There were 70 child tickets and 130 adult tickets sold.
CHECK YOUR SOLUTION USING THE ORIGINAL WORDING
NOT YOUR EQUATIONS..
Sue’s age now is twice Anne’s age. Five years ago Sue’s age was
three times Anne’s age. What is Sue’s age now?
Let S = Sue's age now ; A = Anne's age now
Now S = 2A → 1
5 years ago S − 5 = 3(A − 5)
S − 5 = 3A − 15 → 2
substitute 1 into 2
2A − 5 = 3A − 15
10 = A
1 → S = 20
Sue’s age now is 20 years.
12L of 42% acid solution
Tom’s age now is seven more than half Mary’s age. In seven years
Mary will be five years older than Tom. Find their ages now.
Let T = Tom's age now ; M = Mary's age now
Now T = M
2 +7
Now 2T = M + 14 → 1
In 7 years M + 7 = (T + 7) + 5
M = T+5→ 2
substitute 2 into 1
2T = T + 5 + 14
T = 19
2 → M = 24
Tom is 19 now ; Mary is 24 now.
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