MCR 3U
Date: __________________________
Linear-Quadratic Systems of Equations
In grade 10 you studied the point of intersection between two lines.
We called this the solution to a linear system of equations.
Parallel lines could be distinct (same slope different y-intercept) and have no point of intersection or
coincident (same slope and same y-intercept) and have infinitely many points of intersection.
You determined the point of intersection by graphing, using the substitution method or using the
elimination method. We tended not to use the graphing method because it was time consuming and not
always accurate.
Substitution Method
Steps: 1) Isolate one of the variables in one of the equations.
2) Substitute the expression for the variable you isolated into the second equation.
3) Solve for the variable.
4) Substitute the value into the isolated equation to solve for the second variable.
Ex: Solve 2 y = x + 5 and x − 4 y = 0 using substitution.
Isolate x in the first equation: 2 y − 5 = x
Substitute the expression for x into the second equation: 2 y − 5 − 4 y = 0
Solve for the variable: 2 y − 5 − 4 y = 0
2y − 4y = 5
− 2y = 5
5
y=
−2
−5
y=
or − 2.5
2
Substitute y = -2.5 into 2 y − 5 = x
2(− 2.5) − 5 = x
−5−5 = x
The solution is (-10, -2.5)
− 10 = x
Page 1 of 4.
Elimination Method
Steps:
1) Write equations in the same form, usually a x + b y = c, but standard or y = is okay too!
2) Look for the coefficients of one of the variables to be the same digits (sign does not matter).
If necessary, multiply one or both equations by a whole number to achieve this.
3) a) Same digits, same sign: subtract the two equations.
b) Same digits, opposite sign: add the two equations.
4) Solve the resulting equation.
5) Substitute to find the value of the other variable.
Ex: Solve 2 y = x + 5 and x − 4 y = 0 using elimination.
Write the equations in the same form − x + 2 y = 5
Both equations have a digit of 1 in front of the x, one is positive and the other negative so we
add the equations.
− x + 2y = 5
x − 4y = 0
− 2y = 5
5
y=
−2
−5
y=
or − 2.5
2
Substitute y = -2.5 into x − 4 y = 0
x − 4(− 2.5) = 0
x + 10 = 0
The solution is (-10, -2.5)
x = −10
A linear-quadratic System will have no solution, one solution, or two solutions.
Linear-Quadratic systems
Page 2 of 4.
The same three methods (graphing, substitution, and elimination) can be used to determine the solution to
a linear-quadratic system (or even two quadratic relations).
Ex: Solve y = 2( x − 1) + 3 and y = x + 4
2
I will select substitution since the y is isolated in the first equation.
2( x − 1) + 3 = x + 4 , now solve the resulting quadratic equation
2
2( x − 1)( x − 1) + 3 = x + 4
(
)
2 x 2 − 2x + 1 + 3 = x + 4
2x 2 − 4x + 2 + 3 = x + 4
2x 2 − 4x + 5 = x + 4
2x 2 − 4x − x + 5 − 4 = 0
2 x2 − 5x + 1 = 0
x=
x=
−b ± b 2 − 4ac
2a
5±
(− 5)2 − 4(2)(1)
2(2)
x=
5 ± 25 − 8
4
x=
5 ± 17
4
exact answer
5 − 17
5 + 17
x2 =
4
4
5 − 4.12
5 + 4.12
x1 =&
or x2 =&
4
4
9.12
0.88
x1 =&
x2 =&
4
4
x1 =& 2.28
x2 =& 0.22
x1 =
approximate answer
There are two solutions thus there are two points of intersections. Determine the points by
calculating y; I will use y = x + 4 .
y1 =& 2.28 + 4
y1 =& 6.28
y 2 =& 0.22 + 4
y 2 =& 4.22
The points of intersection are approximately (0.22, 4.22) and (2.28, 6.28).
Assigned work: Pg. 684 - 685 # 1(ace), 2af, 3f and word problems on back
Linear-Quadratic systems
Page 3 of 4.
1) The revenue for a production by a theatre group is y = −50t 2 + 300t , where t is the ticket price in
dollars. The cost for the production is y = 600 − 50t . Determine the ticket price that will allow the
production to break even.
A company breaks even
when revenue = cost
2) a) The graph of y = ( x − 2 ) − 3 is shown on the right.
Draw lines with slope of – 4 that intersect the parabola at
i) one point, ii) two points, and iii) no points.
b) Write the equations of the lines from part a)
c) How are all of the lines with slope – 4 that do not intersect
the parabola related?
2
9
y
8
7
6
5
4
3
2
3) A daredevil jumps off the CN Tower and falls freely for
several seconds before releasing his parachute. His height, h,
in metres, t seconds after jumping can be modelled by:
h = −4.9t 2 + t + 360 before he released his parachute; and
h = −4t + 142 after he released his parachute.
1
-4
-3
-2
-1
-1
x
1
2
3
4
5
6
-2
-3
-4
-5
How long after jumping did the daredevil release his
parachute?
-6
-7
-8
4) A punter kicks a football. Its height, h, in metres, t seconds after the kick is
given by the equation h = −4.9t 2 + 18.24t + 0.8 . The height of an
approaching blocker’s hands is modelled by the equation h = −1.43t + 4.26 ,
using the same time. Can the blocker knock down the punt? If so, at what
point will it happen?
5) Justin is skeet shooting. The height of the skeet
is modelled by the equation h = −5t 2 + 32t + 2 ,
where h represents the height in metres t
seconds after the skeet is released. The path of Justin’s bullet is modelled
by the equation h = 31.5t + 1 , with the same units. How long will it take
for the bullet to hit the skeet? How high off the ground will the skeet be
when it is hit?
6) The height, h, of a baseball, in metres, at time t seconds after it is tossed out of a window is modelled
by h = −5t 2 + 20t + 15 . A boy shoots at the baseball with a paintball gun. The trajectory of the
paintball is given by the equation h = 3t + 3 . Will the paintball hit the baseball? If so, when? At
what height will the baseball be?
7) The revenue for a company producing electronic components is given by y = −20 x 2 − 50 x + 200 ,
where x is the price in dollars of each component. The cost for the production is given by
y = 60 x − 10 . Determine the price that will allow the production to break even.
Linear-Quadratic systems
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