187
Section 11.1 – Solving Systems of Linear Equation Using
Substitution and Elimination
A system of equations consists of two or more equations. A solution to a
system of equations is a point that satisfies all the equations in the system.
In this chapter, our focus will be solving systems of linear equations. If we
are solving a system of two linear equations in two variables, there are
three possible types of systems we could have:
Case #1
Consistent System
The equations are
independent.
Case #2
Inconsistent System
The equations are
Independent.
y-axis
y-axis
4
–2
y-axis
4
2
–4
Case #3
Consistent System
The equations are
dependent.
2
x-axis
2
4
4
–4
–2
2
x-axis
2
4
–4
–2
–2
–2
–2
–4
–4
–4
The solution is the
where the lines
intersect. In this case,
the solution is (2, 1).
The lines have
different slopes.
The lines do not
intersect, so there
is no solution, { }.
The lines are different,
but the lines do have
the same slope.
x-axis
2
4
Every point on the
line is a solution. We
write the solution as:
{(x, y) | ax + by = c}
The lines are the
same or coincident.
To solve a system of equations, we can graph each equation on the same
axes and see where they intersect. The point of intersection will be the
solution to the system. We can then check the answer by plugging in the
solution into both equations and verify that the point makes both of them
true.
Solving the following system by graphing:
{
Ex. 1
€
y=
2
3
x–1
x + 2y = 5
188
Solution:
The first equation is already
solved for y.
1)
2
x–1
3
2
, y-int:
3
y=
m=
(0, – 1)
€ solve the second equation
To
for y, subtract x from both sides
€ divided both sides by 2:
and
2)
x + 2y = 5
1
5
x+
2
2
1
– , y-int:
2
y=–
m=
(0,
5
2
)
€ lines
€ intersect at (3, 1) so the solution is (3, 1). This is a
The
consistent system and the equations are independent.
€
€
You can always check the answer by plugging in the solution into
each equation and see if it satisfies all the equations in the system:
2
x–1
3
2
(3) –
3
y=
1=
x + 2y = 5
1
1 = 1 true
€
Objective #1:
€
(3) + 2(1) = 5
5 = 5 true
So, the solution checks.
Solving Systems of Equations Using Substitution.
Using the substitution method, we will solve one equation for one variable
and substitute that expression for that variable in the other equation. This
will give us one linear equation with one variable. We solve that equation
and that will yield the value for one of the coordinates of our solution. We
take that answer, substitute it into one of the two original equations and
solve to find the value for the other coordinate of our solution.
Solve by substitution:
Ex. 2
€
{
2
3
x–
–
1
x
5
€
€
€
1
5
y=
6
6
1
3
+ y=
3
5
€
€
189
Solution:
It is easiest to clear fractions first. The L.C.D. of equation #1 is 6
and the L.C.D. of equation #2 is 15.
2
3
1)
6
€
2)
€
x–
1
y
6
=
5
6
1
y
6
3
=
5
(multiply both sides by 6)
( 23 x) – 6( ) = 6( 56 )
–
1
x
5
€
(
+
1
1
y
3
)
⇒
4x – y = 5
(multiply both sides by 15)
(1 )
( 3)
15 – x + 15 y = 15
⇒
– 3x + 5y = 9
5
€
€ 5
€ 3
Now, solve equation #1 for y:
4x – €
y = 5 €⇒ €– y = – 4x + 5
⇒
y = 4x – 5
Finally, replace y in equation #2 with 4x – 5:
€ + 5(4x –€5) = 9 €
– 3x
(distribute & combine like terms)
17x – 25 = 9
(solve)
17x = 34
x=2
Since y = 4x – 5, then y = 4(2) – 5 = 8 – 5 = 3.
The solution is (2, 3).
Ex. 3
{
6x – 2y = 4
10y + 5x = 0
Solution:
Solve equation #2 for x:
10y + 5x = 0
⇒ 5x = – 10y ⇒ x = – 2y
Replace x in equation #1 with – 2y:
6(– 2y) – 2y = 4
(multiply and combine like terms)
– 14y = 4
(solve)
y=–
2
7
The solution is
€ #2:
Objective
Since x = – 2y, then x = – 2(–
( 47 , – 27 ).
2
7
)=
4
7
€ Using
€
Solving Systems of Equations
Elimination.
€ Addition
€
A variant of the
Property of Equality that if you have two equations,
you can add the left sides and then add the right sides of the equations to
produce an equivalent equation. In other words, If A = B and C = D, then
A + C = B + D. We will therefore manipulate one or both equations until the
coefficients of one of the variable terms are opposites. This is done by
finding the L.C.M. of the coefficients of the term we are trying to eliminate.
190
We will then add the equations together, eliminating the terms with
opposite coefficients. This will give us one linear equation with one
variable. We solve that equation and that will yield the value for one of the
coordinates of our solution. We then take that answer, substitute it into one
of the two original equations and solve to find the value for the other
coordinate of our solution.
Solve by elimination
{
Ex. 4
6x + 5y = 7
x–y=4
Solution:
We need to manipulate the equations so that the coefficients of one
of the variable terms are opposites. Let’s work on getting rid of y,
the easiest way to do this is to multiply the second equation by 5:
x–y=4 ⇒
5x – 5y = 20
Now, the coefficients of the y-terms are opposites, so add the left
sides and the right sides.
6x + 5y = 7
5x – 5y = 20
11x = 27
(solve)
x=
27
11
Substitute into equation #2:
x–y=4
27
11
€
–y=4
So, y = –
Ex.€5
{
17
.
11
⇒
27
11
–y=
Thus, the solution is
44
11
27
11
(
y=–
⇒
,–
17
11
).
17
11
€
€
3x – 11 = 2(– y +€4)
7(3 – y) = 4(x – 5)
€
€
Solution:
The first step is to simplify the equations and write them in standard
form:
3x – 11 = 2(– y + 4)
7(3 – y) = 4(x – 5)
3x – 11 = – 2y + 8
21 – 7y = 4x – 20
3x + 2y – 11 = 8
4x – 20 = 21 – 7y
1)
3x + 2y = 19
4x + 7y – 20 = 21
2)
4x + 7y = 41
Let’s eliminate the x-terms. The LCM of 3 and 4 is 12.
191
1)
3x + 2y = 19
(multiply by 4)
⇒ 12x + 8y = 76
2)
4x + 7y = 41
(multiply by – 3) ⇒ – 12x – 21y = – 123
Now, the coefficients of the x-terms are opposites, so add the left
sides and the right sides.
12x + 8y = 76
– 12x – 21y = – 123
– 13y = – 47
y=
47
13
To find x, plug y =
47
13
into equation #1:
3x + 2y = 19
€ + 2( 47 ) = 19
3x
3x
13
94
+€
=
13
19
(multiply both sides by 13 to clear fractions)
39x + 94 = 247
€ 39x = 153
x=
€
Hence,
(
€ #3 €
Objective
153
51
=
39
13
51 47
,
is
13 13
)
the solution.
Systems of equations that are inconsistent.
Solve€the€following:
Ex. 6
4x + 2y = – 3
10x + 5y = 2
Solution:
Solve equation #1 for y:
4x + 2y = – 3
⇒
2y = – 4x – 3
⇒
y = – 2x – 1.5
In equation #2, replace y with – 2x – 1.5:
10x + 5y = 2
10x + 5(– 2x – 1.5) = 2
(distribute)
10x – 10x – 7.5 = 2
(combine like terms)
– 7.5 = 2
In the context of solving systems of equations, it means that the lines
are parallel. Hence, the system has no solution.
{
Objective #4
System of dependent equations.
192
Solve the following:
Ex. 7
12x – 6y = – 15
– 4x = – 2y + 5
Solution:
Let’s get rid of the x-terms. We will multiply equation #2 by 3:
– 4x = – 2y + 5 (multiply by 3)
⇒
– 12x = – 6y + 15
Now, the coefficients of the x-terms are opposites, so add the left
sides and the right sides. Be careful to align the equal signs:
12x – 6y =
– 15
– 12x
= – 6y + 15
– 6y = – 6y
(add 6y to both sides)
0=0
The equations are the same.
In this case, we will either solve the equation for x or for y and use
our set builder notation to express the answer:
{ (x, y) | x = 0.5y – 1.25} or { (x, y) | y = 2x + 2.5}
{
Objectives #5 - 7:
System of three linear equations.
The graph of the equation in the form ax + by + cz = d is a plane in three
dimensional space. The points in three-dimensional space have three
coordinates, x, y, and z. We will list them in alphabetical order: (x, y, z).
In solving a system of three linear equations in three variables, we are not
looking at intersecting lines, but intersecting planes. Here are some
possible types of systems we could have:
Type 1:
The solution is a point:
193
Type 2:
The solution is a line:
Two
planes are
the same.
OR
Type 3:
There is no solution:
OR
OR
OR
Two
planes are
the same.
194
Type 4:
Solution is a plane:
All three planes are
the same.
The key to solving a system of three equations with three variables is the
reduce it to a system of two linear equations with two variables. To
accomplish this, pick a variable to eliminate. We pick two equations and
use either elimination or substitution to get rid of that variable. We then pick
a different pair of equations in the system and do the same thing. After it is
reduced to a system of two linear equations in two variables, we then just
solve it like before.
Solve the following:
Ex. 8
1)
x+y+z=3
2)
2x – y + 3z = 16
3)
– x + 2y + 2z = 3
Solution:
Let’s first try to eliminate x. So, take equation #1 plus equation #3:
1)
x+ y+ z=3
3)
– x + 2y + 2z = 3
4)
3y + 3z = 6
We will call this equation #4.
Now, choose a different pair in the system to eliminate x. So, take
equation #2 plus 2•equation #3:
2)
2x – y + 3z = 16
2•3) – 2x + 4y + 4z = 6
5)
3y + 7z = 22
We will call this equation #5
Looking at equation #4 and #5, it is easiest to eliminate the y-terms.
So, take equation #4 plus – 1•equation #5:
4)
3y + 3z = 6
•
– 1 5) – 3y – 7z = – 22
– 4z = – 16
(solve)
z=4
Plug in z = 4 into equation #4 (or #5):
4)
3y + 3(4) = 6
(solve)
3y + 12 = 6
3y = – 6
y=–2
195
Now, plug in y = – 2 & z = 4 in equation #1 (or #2 or #3):
1)
x + (– 2) + (4) = 3
(solve)
x+2=3
x=1
Thus, the solution is (1, – 2, 4).
Ex. 9
1)
x – 3y + 2z = 1
2)
2x + y – z = 3
3)
6x – 4y + 2z = 5
Solution:
First, eliminate z. Take eqn. #1 plus 2•eqn. #2:
1)
x – 3y + 2z = 1
2•2)
4x + 2y – 2z = 6
4)
5x – y = 7
Next, take 2•eqn. #2 plus eqn. #3:
2•2)
4x + 2y – 2z = 6
3)
6x – 4y + 2z = 5
5) 10x – 2y = 11
Looking at equations #4 and #5, let’s get rid of x-terms.
Take – 2•eqn. #4 + eqn. #5:
– 2•4)
– 10x + 2y = – 14
5)
10x – 2y = 11
0=–3
Not possible
This system has no solution.
Ex. 10
1)
2)
5
5
x – 3y + 7z = –
2
2
1
2
2
x+y– z=
3
3
3
3)
2x – y + 4z = – 1
€
Solution:
First clear fractions by multiply both sides of eqn. #1 by 2 and both
sides of eqn. #2 by 3 (we will also list eqn. #3):
2•1)
5x – 6y + 14z = – 5
call this our new equation #1
3•2)
x + 3y – 2z = 2
call this our new equation #2
3)
2x – y + 4z = – 1
Let’s eliminate x. Eqn. #1 + – 5•eqn. #2:
1)
5x – 6y + 14z = – 5
•
– 5 2) – 5x – 15y + 10z = – 10
4)
– 21y + 24z = – 15
196
– 2•Eqn. #2 + eqn. #3:
– 2•2) – 2x – 6y + 4z = – 4
3) 2x – y + 4z = – 1
5)
– 7y + 8z = – 5
Looking at equation #4 and #5, let’s eliminate the y-terms.
Eqn. #4 + – 3•eqn. #5:
4)
– 21y + 24z = – 15
– 3•5)
21y – 24z = 15
0=0
This means that there are an infinite number of solutions. To express
our answer, we will need to take two of the variables and rewrite then
in terms of the third variable. To do this, we will first take equation #5
and solve it for y:
– 7y + 8z = – 5
– 7y = – 8z – 5
y=
8
7
z+
5
7
We will then replace y by
z:
€
8
5
8
7
z+
5
7
and solve for x to get x in terms of
x + 3( z + ) – 2z = 2
(distribute and combine like terms)
7
7
€
10
15
x+
z+
=2
(solve for x)
7
€7
€
10
1
x=–
z–
7
7
€
€
10
1
8
5
Our solution is { (x, y, z) | x = –
z – , y = z + , z is any real
7
7
7
7
€
€
number}.
€
€
Ex. 11
Maylina buys two packages of Skittles and three bottles of
€ four€packages
€ of€Skittles and five bottles
soda for $5.15. Zoë buys
of soda for $9.05. Find the cost of one package of Skittles and one
bottle of soda.
Solution:
Let S = the cost of one package of Skittles
b = the cost of bottle of soda
For Maylina, she buys:
2 Skittles + 3 bottles of soda = $5.15 ⇒ 1) 2S + 3b = 5.15
For Zoë, she buys:
4 Skittles + 5 bottles of soda = $9.05 ⇒ 2) 4S + 5b = 9.05
So, our system is:
197
1)
2S + 3b = 5.15
2)
4S + 5b = 9.05
Let’s multiply equation #1 by – 2 and add:
– 2•1) – 4S – 6b = – 10.30
2)
4S + 5b = 9.05
– b = – 1.25
So, b = 1.25
Substitute 1.25 for b in equation #1 and solve:
2S + 3(1.25) = 5.15
2S + 3.75 = 5.15
2S = 1.40
S = 0.70
So, a package of Skittles costs $0.70 & a bottle of soda costs $1.25.
Ex. 12
Juanita decided to divide $8800 into two investments, one in
bonds that ended up earning 8% annual interest and one in a
mutual fund that ended up yielding 12% annual interest. If the
interest from both was $924, how much did she invest in each?
Solution:
Let b = the amount invested in bonds.
m = the amount invested in a mutual fund
Since there was total $8800 invested, the amount invested in a
mutual fund plus the amount invested in bonds was $8800:
1)
b + m = 8800
The interest earned on the bonds was 8% times the amount
invested in bonds or 0.08b and the interest earned on the mutual
fund was 12% times the amount invested in the mutual fund or
0.12m. The total interest is
2) 0.08b + 0.12m = 924.
So, our system is:
1)
b + m = 8800
2) 0.08b + 0.12m = 924.
Let’s solve equation #1 for m:
b + m = 8800
⇒
m = 8800 – b
Now, replace m is equation #2 by 8800 – b:
0.08b + 0.12(8800 – b) = 924 (distribute & combine like terms)
– 0.04b + 1056 = 924
– 0.04b = – 132
b = 3300
m = 8800 – b = 5500
So, $3300 was invested in bonds and $5500 in a mutual fund.
198
Ex. 13
Find the equation of the quadratic function in the form
y = ax2 + bx + c whose graph passes through the points (– 2, – 29),
(3, – 14), and (1, – 2).
Solution:
The point (– 2, – 29) yields: – 29 = a(– 2)2 + b(– 2) + c or
1)
4a – 2b + c = – 29
The point (3, – 14) yields:
The point (1, – 2) yields:
– 14 = a(3)2 + b(3) + c or
2)
9a + 3b + c = – 14
– 2 = a(1)2 + b(1) + c
3)
a+b+c=–2
or
Thus, we have a system of three equations:
1)
4a – 2b + c = – 29
2)
9a + 3b + c = – 14
3)
a+b+c=–2
Let’s first try to eliminate c. Take equation #1 plus – 1 times equation #3:
1)
4a – 2b + c = – 29
–1•3)
–a–b –c=2
3a – 3b = – 27
(divide both sides by 3)
4)
a–b=–9
We will call this equation #4.
Now, choose a different pair in the system to eliminate c. So, take
equation #2 plus – 1•equation #3:
2)
9a + 3b + c = – 14
–1•3) – a – b – c = 2
5)
8a + 2b = – 12 We will call this equation #5
Looking at equation #4 and #5, it is easiest to eliminate the b-terms.
So, take 2•equation #4 plus equation #5:
2•4)
2a – 2b = – 18
5)
8a + 2b = – 12
10a = – 30
(solve)
a=–3
Plug in a = – 3 into equation #4
4)
–3–b=–9
(solve)
–b=–6
b=6
Now plug in a = – 3, and b = 6 into equation #3 and solve:
–3+6+c=–2
3+c=–2
c=–5
Thus, the quadratic function is y = – 3x2 + 6x – 5.