Download Solving Linear Equations

Solving Linear Equations
An equation that involves only the first power of a single variable and doesn’t involve division by
that variable is called a linear equation in that variable.
Examples: Each of the following is a linear equation in the variable x.
3x = 15 , 2x − 5 = 17 , 3x − 5 = 2x +1, 3(2x − 3) − 5 = 2(x − 5) +16
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A solution (assuming there is one) to a linear equation with one variable is a number, that when
substituted for the variable, results in a true statement.
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Examples:
The equation 3x = 15 has 5 as a solution because 3(5) = 15 .
To check a number to see if it is a solution to a linear equation with one variable, substitute the
number for every occurrence of the variable and see if a true statement results. If it does, then the
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€ number is not a solution.
number is a solution; if it doesn’t, then the
Example: Determine whether 5 is a solution to 3(2x − 3) − 5 = 2(x − 5) +16 .
Substitute 5 for each occurrence of x in the equation and carry out the indicated operations. We get
3(2⋅ 5 − 3) − 5 = 2(5 − 5) +16
3(10 − 3) − 5 = 2(0) +16
3(7) − 5 = 0 +16
21 − 5 = 16
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16 = 16
Because this is a true statement, 5 is a solution to the given equation.
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To solve a linear equation with one variable, you need to get the variable by itself with a
coefficient of 1 on one side of the equation and leave only a single number on the other side.
There are four operations you can use to solve a linear equation with one variable:
1. You can add the same quantity to both sides.
2. You can subtract the same quantity from both sides.
3. You can multiply both sides by the same non-zero quantity.
4. You can divide both sides by the same non-zero quantity.
To solve a linear equation with one variable most efficiently use the following procedure as
demonstrated in the example. (Not every equation will require all of these steps.)
PROCEDURE
EXAMPLE: Solve −4(2x − 3) +1 = 5x − 2(3x + 4)
1. Use the Distribute Property,
−8x +12 +1 = 5x − 6x − 8 Using the Distributive Property to
when applicable, to simplify any
multiply out the parentheses on
additions or subtractions in parenboth sides of the equation.
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theses that are being multiplied.
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2. Combine any like terms on each
−8x +13 = −x − 8
Combining the two constant terms
side of the equation.
on the left and combining the two
x terms on the right.
3. Add or subtract as needed €
to get
all terms that involve the variable
on one side of the equation and all
terms that don’t involve the variable
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on the other side.
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(Thomason – Spring 2015)
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−8x +13 + x = −x − 8 + x
−7x +13 = −8
−7x +13 −13 = −8 −13
−7x = −21
Adding x to both sides.
Simplifying.
Subtracting 13 from both sides.
Simplifying.
p. 1 of 2 4. If the x coefficient is not already
1, divide both sides by the x
coefficient to make it be 1.
−7x −21
=
−7
−7
x=3
So 3 is the solution.
Dividing both sides by –7.
Simplifying.
Note: In step 4 of the procedure,
€ it is sometimes easier to multiply both sides of the equation by the
reciprocal of the x coefficient€than to divide by the coefficient itself. This is especially true if
fractions are involved as in the next example.
Example: To solve 23 x = 95 , the first three steps of the procedure are unnecessary because we
already have only the variable term on one side of the equation and only a constant on the other side.
All that remains to be done is to make the x coefficient be 1. We could divide both sides by the x
coefficient, 23 , but that’s a little complicated. It’s easier to multiply both sides by the reciprocal of
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the x coefficient. So we multiply both sides by 23 as follows:
1 1
1
3⋅ 2 x = 3⋅ 5 !
2 €3
2 9
3
1 1
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and get x = 65 as the solution.
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If you are solving an equation that contains several fractions, it is probably easiest to multiply both
sides by the least common denominator (LCD) of the fractions in the equation. Doing so will “clear
€ the fractions;” that is, change them to integers.
Example: To solve 43 x − 23 = 12 x + 65 , first multiply both sides by 12, the LCD of all the fractions.
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)
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12 43 x − 23 = 12 12 x + 65
4
2
6
€ 2 = 12 1 x + 12 5 !
− 12
1 3
1 2
1 6
1
1
1
1
9x − 8 = 6x +10
3x = 18
x =6
3
( )
12 3 x
1 4
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)
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Distributing on both sides.
Carrying out the indicated multiplications.
Adding 8 to both sides and subtracting 6x from both sides.
Dividing both sides by 3.
If, when attempting to solve a linear equation with one variable, you reach a false statement such as
“0 = 2 ,” it means there is no solution. That is, there is no value of the variable that will make the
given equation true.
Example: Solve 2(3x − 5) = 3(2x −1)
6x −10 = 6x − 3
Distributing on both sides.
6x −10 +10 = 6x − 3 +10
Adding 10 to both sides.
6x = 6x + 7
Simplifying.
6x − 6x€= 6x + 7 − 6x
Subtracting 6x from both sides.
0=7
Simplifying.
Because this statement is never true, there is no solution to the given equation.
If, when attempting to solve a linear equation with one variable, you reach a statement that is always
true, such as “3=3,” it means there are infinitely many solutions. The given equation is true no
matter what number you use for the variable
Example: Solve 2(3x − 5) = 3(2x −1) − 7
6x −10 = 6x − 3 − 7
Distributing on both sides.
6x −10 = 6x −10
Simplifying.
Because this statement is always true, there are infinitely many solutions.
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(Thomason – Spring 2015)
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