Download Solving Linear Equations

Introduction
While it may not be efficient to write out the justification
for each step when solving equations, it is important to
remember that the properties of equality must always
apply in order for an equation to remain balanced.
As equations become more complex, it may be helpful to
refer to the properties of equality used in the previous
lesson.
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2.1.2: Solving Linear Equations
Key Concepts
• When solving equations, first take a look at the
expressions on either side of the equal sign.
• You may need to simplify one or both expressions
before you can solve for the unknown. Sometimes you
may need to combine like terms by using the
associative, commutative, or distributive properties.
• Pay special attention if the same variable appears on
either side of the equal sign.
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2.1.2: Solving Linear Equations
Key Concepts, continued
• Just like with numbers, variables may be added or
subtracted from both sides of the equation without
changing the equality of the statement or the solution
to the problem.
Solving Equations with the Variable in Both Expressions of the Equation
1. Choose which side of the equation you would like the variable to
appear on.
2. Add or subtract the other variable from both sides of the equation using
either the addition or subtraction property of equality.
3. Simplify both expressions.
4. Continue to solve the equation.
5. As with any equation, check that your answer is correct by substituting
the value into the original equation to ensure both expressions are equal.
2.1.2: Solving Linear Equations
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Key Concepts, continued
• Some equations may have no solution. This is the
case when, after you’ve completed all of the
appropriate steps to solve an equation, the result is
something impossible, like 2 = 6. The resulting
equation is never true for any value of the variable.
• Some equations will be true for any value the variable
is replaced with. This is the case when following all of
the appropriate steps for solving an equation results in
the same value on each side of the equal sign, such
as 2x = 2x. The resulting equation is always true for
any value of the variable.
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2.1.2: Solving Linear Equations
Key Concepts, continued
• Other equations will only have one solution, where the
final step in solving results in the variable equal to a
number, such as x = 5.
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2.1.2: Solving Linear Equations
Common Errors/Misconceptions
• performing the wrong operation when isolating the
variable
• incorrectly combining terms
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2.1.2: Solving Linear Equations
Guided Practice
Example 1
Solve the equation 5x + 9 = 2x – 36.
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2.1.2: Solving Linear Equations
Guided Practice: Example 1, continued
1. Move the variable to one side of the
equation.
Notice that the same variable, x, is on both sides of
the equation: 5x is on the left of the equation and 2x is
on the right. It makes no difference whether you
choose to have the variables on the left or on the right;
your solution will remain the same. It’s common to
have the variable on the left, but not necessary.
It’s often easier to move the variable with the smallest
coefficient to the opposite side of the equation. Here,
2x is smaller than 5x, so let’s move 2x.
2.1.2: Solving Linear Equations
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Guided Practice: Example 1, continued
2x is positive, so to get it to the other side of the
equal sign you will need to subtract it from both
expressions in the equation.
It helps to line up what you are subtracting with the
terms that are similar in order to stay organized. In
this case, we are subtracting variables from
variables.
5x + 9 = 2x - 36
-2x
- 2x
3x + 9 =
- 36
2.1.2: Solving Linear Equations
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Guided Practice: Example 1, continued
When 2x is subtracted, it’s important not to forget the
remaining terms of each expression. Look out for
subtraction signs that now act as negative signs.
Here, since 36 was originally being subtracted from
2x, the subtraction sign left behind makes 36
negative.
3x + 9 = –36
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2.1.2: Solving Linear Equations
Guided Practice: Example 1, continued
2. Continue to solve the equation 3x + 9 = –
36.
To isolate x, subtract 9 from both expressions in the
equation.
3x + 9 = -36
-9
3x
-9
= -45
Divide
3x both
-45 expressions by the coefficient of x, 3.
=
3
3
x = -15
2.1.2: Solving Linear Equations
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Guided Practice: Example 1, continued
3. The solution to the equation 5x + 9 = 2x –
36 is x = –15.
A quick check will verify this. Substitute –15 for all
instances of x in the original equation, and then
evaluate each expression.
5x + 9 = 2x – 36
5(–15) + 9 = 2(–15) – 36
–75 + 9 = –30 – 36
–66 = –66
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2.1.2: Solving Linear Equations
Guided Practice: Example 1, continued
Our check verified that both sides of the equation are
still equal; therefore, x = –15 is correct.
✔
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2.1.2: Solving Linear Equations
Guided Practice: Example 1, continued
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2.1.2: Solving Linear Equations
Guided Practice
Example 5
1
Solve the literal equation A = (b1 + b2 )h for b1.
2
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2.1.2: Solving Linear Equations
Guided Practice: Example 5, continued
1. Isolate b1.
As we saw in Unit 1, to solve literal equations for a
specific variable, we follow the same steps as solving
equations.
In this equation, we could distribute
1
over b1 + b2,
2
but this may cause more work for us. Instead, let’s
get rid of the fraction. Multiply both sides of the
1
equation by the inverse of , or 2.
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2.1.2: Solving Linear Equations
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Guided Practice: Example 5, continued
1
A = (b1 + b2 )h
2
1
2· A = 2· (b1 + b2 )h
2
2A = (b1 + b2 )h
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2.1.2: Solving Linear Equations
Guided Practice: Example 5, continued
2. Again, you could distribute h over b1 + b2,
but it’s more efficient to divide both sides
of the equation by h.
2A = (b1 + b2 )h
2A (b1 + b2 )h
=
h
h
2A
= b1 + b2
h
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2.1.2: Solving Linear Equations
Guided Practice: Example 5, continued
3. Finally, to solve the equation for b1,
subtract b2 from both expressions of the
equation.
2A
h
-b2
2A
h
= b1 + b2
- b2
- b2 = b1
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2.1.2: Solving Linear Equations
Guided Practice: Example 5, continued
1
4. The equation A = (b1 + b2 )h solved for b1
2
2A
is
- b2 = b1.
h
2A
The equation can be rewritten as b =
- b2 using
1
h
the symmetric property of equality.
✔
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2.1.2: Solving Linear Equations
Guided Practice: Example 5, continued
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2.1.2: Solving Linear Equations