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Chapter 1 Solving Linear Equations
Section 1.5 Formulas Section 1.6 Absolute Value Equations
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Section 1.5 Formulas Objective: Solve linear formulas for a given variable. When solving formulas for a variable we need to focus on the one variable we are trying to solve for, all the others are treated just like numbers. For example: Solve the formula π+ππ= π π for π ) for π ) for π
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Solution Solve π+ππ= π π for π π= π 3 β2π 2) for π
2π= π 3 βπ π= π 3 βπ 2 3) for π π+2π β3=π 3π+6π=π In your solution, the variable you solve for should be isolated on one side, and on the other side it doesnβt have this variable anymore. For example, in part 1), a is isolated on the left side, and no a on the right side.
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Isolate the term containing the variable that we want to solve for
Example 1: Solve the following formula for π₯ π¦ = 5ππ₯β6π π₯ = Solution: π¦+6π=5ππ₯ π¦+6π 5π =π₯
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Example 2: Solve the formula for π. π΄ = 12 β (π + π) π =
Multiply out first and isolate the term containing the variable that we want to solve for Solution: π΄=12βπ+12βπ π΄β12βπ=12βπ π΄β12βπ 12β =π Note: Cases are sensitive when you write variables. That means, you cannot enter upper case A as lower case a. They represent two different variables. Example 2: Solve the formula for π. π΄ = 12 β (π + π) π =
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When there are more than one term containing the variable that we want to solve for, collect those terms on one side and factor out the variable then solve. Example 3: The formula, π΄ = 2ππ€ + 2πβ + 2π€β, gives the surface area π΄, of a rectangular solid with length, width, and height, π, π€, and β, respectively. Solve the formula for π€. π΄ = 2ππ€ + 2πβ + 2π€β π€ = Solution: π΄β2πβ=2ππ€+2π€β π΄β2πβ=π€ 2π+2β π΄β2πβ 2π+2β =π€
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Solve the equation for h: 14(πββ)=βπ’+2π
Multiply out first; Collect the terms containing the variable that we want to solve for; Factor out the variable and solve. Example 4: Solve the equation for h: 14(πββ)=βπ’+2π β = Solution: 14πβ14β=βπ’+2π β14βββπ’=2πβ14π β β14βπ’ =2πβ14π β= 2πβ14π β14βπ’
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Fractions---Multiply by LCD first
Example 5: Solve the equation for p. 5 π = 8 π‘ + 3π π = Solution: πΏπΆπ· π , π‘ =π π‘ Multiply π π‘ on both sides, or by every term. 5 π π π‘ = 8 π‘ π π‘ + 3π π π‘ 5π‘=8π +3ππ π‘ 5π‘β8π =3ππ π‘ 5π‘β8π 3π π‘ =π
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More Practice Solve the following formula for π π=πππ‘ π=
Solve the formula for the specified variable. π=2ππ‘β+2ππβ β=
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More Practice Solve for π§ 8π§β8π=πβ6 π§ = Solve the formula for π.
π΄=12β(π+π) π=
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More Practice Solve the formula for π . π π‘+π€π=π π+π π = Solve for π
π€=45(π¦π+π) π =
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More Practice Solve the formula for π 6π+ππ π =π π=
Solve the equation for π: 1 4 (π¦βπ)=ππ+4π π=
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Section 1.6 Absolute Value Equations
Objective: Solve linear absolute value equations. The absolute value is the distance between a number and zero on a number line. So an absolute value should not be a negative number. It doesnβt make sense to say a distance is negative. If π =π, then π₯=4 ππ β4.
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Example 1: Solve the equation. If there is more than one solution, separate them with a comma. |π₯|=6 π₯=6 ππ β6
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You should isolate the absolute value part first.
Example 2: Solve the equation. If there is more than one solution, separate them with a comma. 8β|π₯|=4 π₯= Solution: β π₯ =4β8 β π₯ =β4 π₯ = β4 β1 π₯ =4 π₯=4 ππ β4
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More Practice Solve the equation. If there is more than one solution, separate them with a comma. |π₯|=2 Solve the equation. If there is more than one solution, separate them with a comma. 4β|π₯|=2
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Consider the expression inside the absolute value as a whole, write into two equations and then solve for the variable. Solution: 5π₯+1= π₯=20β1 5π₯=19 π₯= 19 5 Example 3: Solve the equation. If there is more than one solution, separate them with a comma. |5π₯+1|=20 π₯= ππ 5π₯+1=β20 5π₯=β20β1 5π₯=β21 π₯=β 21 5 There are two different solutions. They are not just opposite to each other.
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Example: Solve the equation. If there is more than one solution, separate them with a comma. 4π₯+3 =β11 π₯= Solution: Since an absolute value should not be a negative number. This problem has no solution.
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More Practice Solve the equation |5π₯β1|=19 The solutions are:
|2π₯β1|=18 The solutions are:
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Remember you should isolate the absolute value part first, before you apply the definition and write into two equations. Example 5: Solve the equation. If there is more than one solution, separate them with a comma. 2β4|5π₯β4|=β22 π₯= Solution: β4 5π₯β4 =β22β2 β4 5π₯β4 =β24 5π₯β4 = β24 β4 5π₯β4 =6 5π₯β4=6 5π₯=6+4 5π₯=10 π₯= π₯=2 Or 5π₯β4=β6 5π₯=β6+4 5π₯=β2 π₯=β 2 5
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More Practice Solve the equation |4π₯+4|=13 The solutions are:
β5+4|4π₯β1|=11 The solutions are:
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More Practice Solve the equation 5β2|5π₯+3|=β11 The solutions are:
5β4|4π₯+3|=β3 The solutions are:
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Example 6: Solve the equation
Example 6: Solve the equation. If there is more than one solution, separate them with a comma. |4π₯β1|=|7π₯β4| π₯= Solution: Both sides contain only absolute values, we can start writing as two equations, and solve each equation for x. 4π₯β1=7π₯β4 4π₯β7π₯=β4+1 β3π₯=β3 π₯=1 Or 4π₯β1=β 7π₯β4 4π₯β1=β7π₯+4 4π₯+7π₯=4+1 11π₯=5 π₯= 5 11
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More Practice Solve the equation |2π₯+1|=|4π₯β2| The solutions are:
|2π₯+3|=|5π₯β1| The solutions are:
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