Download 9 Solving Systems of Equations

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Solving Systems of Equations
Solving Linear Systems of Equations
UNDERSTAND Recall that a system of equations is a set of two or more equations with
the same variables. The solution to a system of linear equations is the set of values of the
variables that make all equations in the system true. For a system with 2 variables, the
solutions are also the points on the coordinate plane where the graphs of the two equations
intersect.
y
The graph on the right shows y 5 22x 2 1 and y 5 4x 1 5.
The graphs intersect at the point (21, 1). The point gives
the x- and y-values of the solution. Test to see that the point
makes both equations true.
y 5 22x 2 1
y 5 4x 1 5
1 5 22(21) 2 1
1 5 4(21) 1 5
15221
1 5 24 1 5
151
151
6
4
2
–6
–4
–2
0
2
4
6
x
–2
y 4x 5
–4
y 2x 1
–6
UNDERSTAND A linear system can have no solution, one solution, or infinitely many
solutions. If the graphs do not intersect at all, the system has no real solution and is said to
be inconsistent. If an equation is a constant multiple of another, the system has infinitely
many solutions.
Use elimination to solve the system above. Eliminate y by subtracting the second equation
from the first.
y 5 22x 2 1
2(y 5 4x 1 5) Remember to distribute the negative sign.
0 5 26x 2 6
6x 5 26
x 5 21
The value of x is the same as the one found on the graph. To find the corresponding value of y,
substitute the value of x into one of the original equations.
y 5 22(21) 2 1 5 1
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UNDERSTAND A system can also be solved algebraically using substitution or elimination.
To use elimination, add or subtract the equations to write an equation in one variable.
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Connect
Solve the system using elimination and by graphing.
3x 1 2y 5 4
4x 2 3y 5 11
1
Solve the system using elimination.
Choose a variable to eliminate. Eliminate y.
3(3x 1 2y 5 4)
9x 1 6y 5 12
2(4x 2 3y 5 11)
8x 2 6y 5 22
Add the two equations to eliminate the
y term.
9x 1 6y 5 12
1 (8x 2 6y 5 22)
17x 5 34
2
Solve the system by graphing.
x 5 2
Draw both graphs on the same coordinate
plane. Identify the point of intersection.
Use one of the original equations to solve
for y.
y
3x 1 2y 5 4
3(2) 1 2y 5 4
6
2y 5 22
4x 3y 11
y 5 21
3x 2y 4
4
2
–6
–4
–2
0
–2
2
4
(2, 1)
6
x
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–4
–6
3
Identify and test the solution.
Using elimination, the solution is (2, 21).
The point of intersection of the graphs
is (2, 21). Substitute the point into each
equation to test.
3x 1 2y 5 4: 3(2) 1 2(21) 5 4 
4x 2 3y 5 11: 4(2) 2 3(21) 5 11 
▸ The solution to the system is (3, 22).
TRY
Solve the system using elimination and
by graphing.
3y 2 6x 5 12
y 2 2x 5 4
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Solving a Linear-Quadratic System of Equations
UNDERSTAND In a system of equations, the equations may not all be linear. Another type
of system is a linear-quadratic system. In this system one equation is linear and the other is
quadratic. Recall that the graph of a quadratic equation is a special curve called a parabola.
A linear-quadratic system can be solved by graphing or using substitution.
The graph on the right shows y 5 x2 and y 5 2x.
The graphs have 2 points of intersection: (0, 0) and (2, 4).
Test to see that they make both equations true.
(0, 0)
x 5 0 and y 5 0
2
y
10
y x2
8
y 2x
(2, 4)
6
x 5 2 and y 5 4
4
2
0 50
2 54
0 5 2(0) 
4 5 2(2) 
(2, 4)
2
–6
–4
–2
0
2
4
6
x
–2
UNDERSTAND A line can intersect a parabola at two points, one point, or no points. So, a
system of linear quadratic equations can have two solutions, one solution, or no solutions.
UNDERSTAND A linear-quadratic system can also be solved algebraically using substitution.
Use substitution to solve the system above. Substitute 2x for y in the quadratic equation to get
an equation in one variable.
2x 5 x2
0 5 x2 2 2x
Solve the equation by factoring.
x 5 0 and x 5 2
The values of x are the same as the ones found on the graph. To find the corresponding values
of y, substitute the values of x into one of the original equations.
y 5 x2 y 5 x2
y 5 02
y 5 22
y 5 0
y54
(0, 0) and (2, 4) are solutions to the system. These solutions are the same as the ones found on
the graph.
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0 5 x(x 2 2)
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Connect
Solve the system using substitution and by graphing.
y 5 x2 2 2x 2 3
y 5 2x 2 1
1
Solve the system using substitution.
Substitute 2x 2 1 for y in the quadratic
equation. Then solve for x.
y 5 x2 2 2x 2 3
2
2x 2 1 5 x 2 2x 2 3
0 5 x2 2 x 2 2
0 5 (x 1 1)(x 2 2)
2
Solve the system by graphing.
x 5 21 and x 5 2
Use the linear equation to solve for y.
x 5 21: y 5 2(21) 2 1 5 0
Draw both graphs on the same coordinate
plane. Identify any points of intersection.
y
x 5 2: y 5 2(2) 2 1 5 23
6
4
2
(1, 0)
–6
–4
0
–2
2
4
6
x
–2
y x 2 2x 3
(2, 3)
–4
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–6
y x 1
3
Identify the solution.
Using substitution, the solutions are (21, 0)
and (2, 23).
The graphs intersect at two points: (21, 0)
and (2, 23).
▸ The solutions to the system are (21, 0)
and (2, 23).
CHECK
Substitute both solution points into each
equation to check that they are solutions.
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EXAMPLE A Solve the system using substitution and by graphing.
x 2 2y 5 7
4x 1 y 5 10
1
Solve the system using substitution.
Solve the first equation for x: x 5 2y 1 7
Substitute 2y 1 7 for x in the second
equation. Then solve for y.
4(2y 1 7) 1 y 5 10
8y 1 28 1 y 5 10
9y 5 218
y 5 22
Use the first equation to solve for x.
2
Solve the system by graphing.
x 2 2y 5 7
x 2 2(22) 5 7
x1457
Draw both graphs on the same coordinate
plane. Identify the point of intersection.
y
x53
6
4x y 10
4
2
–4
–2
0
–2
x 2y 7
2
4
6
x
(3, 2)
–4
–6
3
Identify the solution.
Using elimination, the solution is (3, 22).
The point of intersection of the graphs is
(3, 22).
▸ The solution to the system is (3, 22).
SC U S S
DI
Which algebraic method would you use
to solve the system? Explain.
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–6
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EXAMPLE B Solve the system.
x2 1 y2 5 10
y 5 2x 1 2
1
Solve by substitution.
Substitute 2x 1 2 for y in the first equation.
x2 1 y2 5 10
x2 1 (2x 1 2)2 5 10
Simplify and solve for x.
x2 1 x2 2 4x 1 4 5 10
2
Graph both equations on your graphing
calculator, adjusting the window as
needed.
2x2 2 4x 2 6 5 0
2(x2 2 2x 1 3) 5 0
(x 1 1)(x 2 3) 5 0
x 5 21 and x 5 3
Use the linear equation to solve for y.
x 5 21: y 5 2(21) 1 2 5 3
x 5 3: y 5 2(3) 1 2 5 21
There appear to be 2 points of intersection.
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Hit 2nd TRACE to open the CALC menu,
and select option 5: intersect. Move the
cursor to one of the points of intersection
and hit ENTER 3 times. The screen shows
that the intersection point is (21, 3).
Repeat the process for the other
intersection point, (3, 21).
3
Identify the solutions.
Using substitution, the solutions are (21, 3)
and (3, 21).
The points of intersection of the graphs are
(21, 3) and (3, 21).
▸ The solutions to the equations are
(21, 3) and (3, 21).
SC U S S
DI
How many different solutions can there
be for a system composed of a circle and
a line? Explain.
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Practice
Use the graphs to solve the system.
REMEMBER Intersections of the graphs
represent solutions to the system.
1.
2.
y
–6
–4
–2
y
6
6
4
4
2
2
0
2
4
6
x
–6
–4
–2
0
–2
–2
–4
–4
–6
–6
2
4
6
x
Solve the system using algebra.
3.2x 1 5y 5 24
4. HI
5.
Look for a way to eliminate
one of the variables.
y 5 x2 1 x 2 1
y5x
84 6. y 5 x2 1 2x 2 8
y5x26
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y 1 x 5 23
NT
3x 2 y 5 11
y 2 3x 5 4
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Solve each system.
7.
x 1 2y 5 10
8. 3x 2 y 5 2
9. 4x 2 2y 5 9
y 5 x2
y 5 2x 1 3
10. y 5 x2 2 5x 1 6
3x 2 y 5 5
2y 5 x 2 2
Solve.
11.
EXTEND On the same coordinate plane, graph the following three equations.
y
x2 1 y2 5 16
x1y54
6
4
y 5 2x 1 4
2
Use the graph to solve each system.
–6
–4
–2
0
2
4
6
x
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–2
–4
–6
a.
x1y54
b. y 5 2x 1 4
c.
2
2
x 1 y 5 16
y 5 2x 1 4
x2 1 y2 5 16
x1y54
d. 2
2
x 1 y 5 16
x1y54
y 5 2x 1 4
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