Download Solve Systems of Equations by the Substitution Method

Solve Systems of Equations by the Substitution Method
Solving a system of equations by graphing has several limitations. First,
it requires the graph to be perfectly drawn – if the lines are not straight,
we may arrive at the wrong answer. Second, graphing is not a great
method to use if the answer is really large or if the answer is a decimal
or fraction. For these reasons, graphing is rarely used to solve systems.
The first algebraic approach, discussed in this section, is called the
substitution method. When the value of one variable is known, it can be
substituted into the other equation.
Example 1: Solve the system: x = 5
y = 2x – 3
y = 2(5) – 3
y = 10 – 3
y=7
Substitute 5 for x into the second equation
The solution to this system is (5, 7)
When one variable equals an expression, that expression can also be
substituted into the other equation. It is important to use parenthesis
around the substituted expression to lessen the chance of making a
mistake when distributing. Once that equation is solved and a value is
known for one of the variable, we proceed as in Example 1.
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 2: Solve the system: 2x – 3y = 7
y = 3x – 7
2x – 3(3x – 7) = 7
2x – 9x + 21 = 7
- 7x + 21 = 7
- 21 - 21
- 7x = -14
-7
-7
x=2
y = 3(2) – 7
y=6–7
y = -1
Substitute (3x – 7) into the first equation
Distribute -3
Combine like terms
Subtract 21 from both sides
Divide both sides by -7
Substitute 2 into the second equation
The solution to this system is (2, -1)
To check the answer: 2(2) – 3(-1) = 7
4+3=7
7=7
True
-1 = 3(2) – 7
-1 = 6 – 7
-1 = -1
True
Since (2, -1) satisfies both equations, it is the solution to the system
Sometimes the lone variable (variable without a coefficient) is not alone
on one side of the equation. In this case, we isolate it by solving for the
lone variable.
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 3: Solve the system: 3x + 2y = 1
x – 5y = 6
x – 5y = 6
+ 5y + 5y
x = 6 + 5y
3(6 + 5y) + 2y = 1
18 + 15y + 2y = 1
18 + 17y = 1
- 18
- 18
17y = -17
17
17
y = -1
Add 5y to both sides to isolate x
Substitute (6 + 5y) into the first equation
Distribute 3
Combine like terms
Subtract 18 from both sides
x = 6 + 5(-1) Substitute -1 into the second equation
x=6–5
x=1
The solution to this system is (1, -1)
To check the answer: 3(1) + 2(-1) = 1
3–2=1
1=1
True
1 – 5(-1) = 6
1+5=6
6=6
True
Since (1, -1) satisfies both equations, it is the solution to the system
Just as with graphing, it is possible to have no solution (parallel lines) or
infinite solutions (same line) when solving algebraically.
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 4: Solve the system: y = 3x – 4
2y – 6x = -8
2(3x – 4) – 6x = -8 Substitute (3x + 4) into the second equation
6x – 8 – 6x = -8
-8 = -8 Combine like term & the variables are gone
This is a true statement, so there are infinite solutions,
meaning the two equations are equivalent and represent the same line.
Example 5: Solve the system: 6x – 3y = -9
y = 2x + 5
6x – 3(2x + 5) = -9 Substitute (2x + 5) into the first equation
6x – 6x – 15 = -9
-15 = -9 Combine like terms & the variables are gone
This is a false statement, so there is no solution,
meaning the two equation represent two different parallel lines.
If there is no lone variable, we select one variable in one of the equations
to isolate. In this situation, it is a good possibility that the resulting
equation will contain fractions.
Example 6: Solve the system: 5x – 6y = -14
-3x + 4y = 12
-3x + 4y = 12
+3x
+3x
4y = 12 + 3x
4 4 4
Add 3x to both sides
Divide each term by 4
3
y=3+ x
4
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
3
3
5x – 6(3 + x) = -14
Substitute (3 + x) into the first equation
5x – 18 –
x = -14
Distribute 6
x = -14)
Clear the fraction by multiplying by LCD
2(5x – 18 –
4
9
2
9
2
10x – 36 – 9x = -28
x – 36 = -28
+36 +36
x
=8
-3(8) + 4y = 12
-24 + 4y = 12
+24
+24
4y = 36
4 4
y=9
4
Add 36 to both sides
Substitute 8 into the second equation
Add 24 to both sides
Divide both sides by 4
The solution to this system is (8, 9)
To check the answer: 5(8) – 6(9) = -14
40 – 54 = -14
-14 = -14
True
-3(8) + 4(9) = 12
-24 + 36 = 12
12 = 12
True
Since (8, 9) satisfies both equations, it is the solution to the system
When two equations equal the same value, they must be equivalent to
each other. Basically we set the equations equal to each other and solve
for the remaining variable. Then we can substitute the resulting into
either equation.
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
3
Example 7: Solve the system: y = x + 2
4
2
y= x+1
3
3
2
4
3
3
2
4
3
x+2= x+1
Both equations equal y, so they equal each other
12( x + 2 = x + 1) Multiply by the LCD to clear the fractions
9x + 24 = 8x + 12
- 8x
- 8x
x + 24 = 12
- 24 - 24
x
= -12
3
Subtract 8x from both sides
Subtract 24 from both sides
y = (-12) + 2
Substitute -12 into the first equation
y = -9 + 2
y = -7
The solution to this system is (-12, -7)
4
3
To check the answer: -7 = (-12) + 2
4
-7 = -9 +2
-7 = -7
True
2
-7 = (-12) + 1
3
-7 = -8 + 1
-7 = -7
True
Since (-12, -7) satisfies both equations, it is the solution to the system
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)