Download 5.b Part 1 Substitution and 6.a Part 2 Applications

Solving Linear Systems
Algebraically
Solving Linear Systems
Algebraically
There are two methods of solving a
system of equations algebraically:
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Substitution (today)
Elimination (next day)
Substitution
Substitution
1) Isolate either x or y in one of the equations. (the easier
one you decide)
2) Substitute the expression for whichever variable you
isolated (i.e. x or y) into the other equation.
3) Solve for the missing variable.
4) Use the value of the variable you just found to find the
other variable by substituting back into either equation.
5) State the point of intersection.
Substitution
● Example 1
Solve the system: x - 2y = -5 (i)
y=x+2
(ii)
Notice: One equation is already solved for one variable.
Plug in (x + 2) for y in the first equation.
x - 2y = -5
x - 2(x + 2) = -5
● We now have one equation with one variable. Simplify and solve.
x - 2x – 4 = -5
-x - 4 = -5
-x = -1
x=1
● Substitute 1 for x in either equation to find y.
y=x+2
1+2=3
● The solution is (1, 3).
Substitution
● Let’s check the solution. The answer (1, 3) must check
in both equations.
x - 2y = -5
y=x+2
1 - 2(3) = -5
3=1+2
-5 = -5
3=3
Example 2 Solve the system:
2x + 4y = 10 (i) and
3x – y = – 6 (ii) using substitution.
a) Isolate y in: 3x – y = – 6
(ii)
– y = – 3x – 6
y = 3x + 6
(b) Substitute y = 3x + 6 into (i)
2x + 4(3x + 6) = 10 (i)
solve ….
2x + 4y = 10 (i)
2x + 4(3x + 6) = 10 (i)
2x + 12x + 24 = 10
14x + 24 = 10
14x = 10 – 24
14x = – 14
x=–1
Substitute x = – 1 into (i)
3x – y = – 6 (ii)
2(– 1) + 4y = 10
–2 + 4y = 10
4y = 10 + 2
4y = 12
y=3
(–1, 3)
Example 3: Solve the system of equations
x + 2y = 4 (i) and 2x + 3y = 7 (ii)
Isolate x in (i)
x = 4 – 2y
Sub in (ii)
2(4 – 2y) + 3y = 7
8 – 4y + 3y = 7
8–y =7
–y =7–8
–y =–1
y = 1
Sub y = 1 into (i)
Ex: Solve the system of equations
x + 2y = 4 (i) and 2x + 3y = 7 (ii)
Sub y = 1 into (i)
x + 2(1) = 4
x+2=4
x=4–2
x=2
Solution
(2, 1)
Which would you solve for?
Solve the systems by substitution:
1.
x=4
2x - 3y = -19
2.
3x + y = 7
4x + 2y = 16
3.
2x + y = 5
3x – 3y = 3
4.
2x + 2y = 4
x – 2y = 0
Extra Example Substitution
2x + 4y = 18
(i) and
y = 3x – 13 (ii) using substitution.
Substitute y = 3x –13 into (i)
2x + 4(3x –13 ) = 18
(i)
2x + 12x – 52 = 18
14x = 18 + 52
14x = 70
x =5
Continued
2x + 4y = 18
y = 3x – 13
(i)
(ii)
Substitute x = 5 into (i)
2(5) + 4y = 18 (i)
10 + 4y = 18
4y = 18 – 10
4y = 8
y=2
Solution
(5, 2)
Part 2 Applications
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
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Declare variables
Write 2 equations
Solve the system for the variables
(A check is a good idea)
Make a concluding statement (with
units)
Example
Ace Heating charges $40 service plus $25/h for labour.
Bill’s Repairs charges $20 service plus $30/h.
Which place is best to use (explain)?
h be the time of the repair (in hours)
Let…
C be the total cost of the repair (in $)
Let…
C = 25h + 40 (i)
C = 30h + 20 (ii)
Sub C = 30h+20 into (i)
30h + 20 = 25h + 40
5h = 20
h=4
Sub h = 4 in (1)
C = 25(4) + 40
C = 100 + 40
C = 140
The two company’s will charge the same cost at a repair that takes 4 hours
It is cheaper to go with Bill’s for a repair that is expected to take less than 4
hours. (as seen from the smaller start-up cost)
It is cheaper to go with Ace for a repair that is expected to take more than 4
hours.