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When solving real-life problems, it is usually possible to write equations in more than one variable. Sometimes solving the equation for the variable of interest is needed. In this lesson, equations involving two or more variables will be solved for a particular variable.
### Catch-Up and Review

**Here is some recommended reading before getting started with this lesson.**

Challenge

Consider $△ABC$ with side lengths $a,$ $b,$ and $c.$

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b If $P,$ $a,$ and $b$ are known, write an equation by solving the formula for $c.$

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Challenge

Maya plans to jog $150$ meters at a constant speed of $2.5$ meters per second.

The distance traveled $d$ is found using the following equation.$d=v⋅t $

In this equation, $v$ is the speed and $t$ is the time. a Determine how long it takes Maya to jog this distance by trying some values for $t$ until a true statement is obtained.

b Find a different way to solve the question.

Discussion

Two equations are called equivalent equations if they have the same solution. Equations are often solved by applying the Properties of Equality. Each time a property is applied, an equivalent equation is produced. Consider the following equation.
Here, two equations that are equivalent to the given equation were created as a result of applying the Subtraction Property of Equality. They are equivalent because $15$ is the solution to *all* these equations.

$y+3=18 $

To solve this equation, $3$ must be subtracted from both sides.
$y+3=18$

SubEqn

$LHS−3=RHS−3$

$y+3−3=18−3$

Simplify left-hand side

$y+3 −3 =18−3$

SubTerms

Subtract terms

$y=15$

$I.II.III. y+3=18y+3−3=18−3y=15 $

The obtained solution can be checked by substituting $y=15$ in the original equation. If a true statement results, the solution is correct.
A true statement was obtained. Therefore, $y=15$ is indeed a solution to the original equation.Example

Dominika is finishing up her homework on Friday so she can enjoy the weekend. She is told that three of the following equations are equivalent equations. ### Hint

### Solution

### Equation I

In the first equation, $11$ is being added to $t.$ Since the inverse operation of addition is subtraction, applying the Subtraction Property of Equality would simplify the equation.
The solution to the first equation is $t=9.$ ### Equation II

To isolate the variable, the Subtraction Property of Equality will be applied first.
To remove the negative sign from the variable $t,$ each side of the equation will be divided by $-1.$ In other words, the Division Property of Equality will be applied.
The solution to Equation II is $t=-9.$ ### Equation III

To solve Equation III for $t,$ $7$ should be subtracted from both sides of the equation.
To remove the fraction, the equation needs to be multiplied by $3.$ To do so, apply the Multiplication Property of Equality.
The solution to Equation III is $t=9,$ which is the same as the solution to Equation I. Therefore, the first and third equations are equivalent. ### Equation IV

In this equation, $40$ is being subtracted from $5t.$ Since the inverse operation of subtraction is addition, applying the Addition Property of Equality will simplify the equation.
By applying the Division Property of Equality, the equation can be further simplified.
The last equation has the same solution as Equations I and III. This means that Equations I, III, and IV are equivalent equations.

$EquationsI:II:III:IV: t+11=207−t=163t +7=105=5t−40 $

Help Dominika identify the equation that is not equivalent to the other three. {"type":"choice","form":{"alts":["Equation I","Equation II","Equation III","Equation IV"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}

Equivalent equations have the same solution(s). Use the Properties of Equality and inverse operations to solve for the variable $t.$

Each equation will be solved for $t,$ and then the solutions will be compared.

$7−t=16$

SubEqn

$LHS−7=RHS−7$

$7−t−7=16−7$

CommutativePropAdd

Commutative Property of Addition

$7−7−t=16−7$

SubTerms

Subtract terms

$-t=9$

$-t=9$

DivEqn

$LHS/(-1)=RHS/(-1)$

$-1-t =-19 $

▼

Solve for $t$

DivNegNeg

$-b-a =ba $

$1t =-19 $

MoveNegDenomToFrac

Put minus sign in front of fraction

$1t =-19 $

DivByOne

$1a =a$

$t=-9$

$45=5t$

DivEqn

$LHS/5=RHS/5$

$545 =55t $

CalcQuot

Calculate quotient

$9=55t $

MovePartNumRight

$ca⋅b =ca ⋅b$

$9=55 t$

QuotOne

$aa =1$

$9=1t$

IdPropMult

Identity Property of Multiplication

$9=t$

RearrangeEqn

Rearrange equation

$t=9$

$Equaivalent EquationsI,III,andIV Solution9 $

Therefore, Equation II is not equivalent to any of those equations.
Discussion

A literal equation is an equation that is comprised mostly or entirely of variables. The area of a triangle, for example, is expressed by the following literal equation.

$A=21 bh $

Here, $b$ represents the length of the base and $h$ represents the height of the triangle. Therefore, formulas can be seen as literal equations. Using the Properties of Equality, a literal equation can be solved for a variable, which means isolating the variable on one side of the equation.Example

While at the planetarium on Saturday, Dominika learns about Annie Jump Cannon, known as the census taker of the sky,

who introduced the *Harvard Spectral Classification*. In its simplest form, this system classifies stars according to their surface temperatures.

Stellar Classification | ||
---|---|---|

Class | Temperature (K) | Apparent Color |

O | $≥30000$ | blue |

B | $10000−30000$ | blue white |

A | $7500−10000$ | white |

F | $6000−7500$ | yellow white |

G | $5000−6000$ | yellow |

K | $3500−5000$ | light orange |

M | $2000−3500$ | orange red |

a To convert temperatures in degrees Celsius $C$ to temperatures in Kelvin $K,$ the following literal equation is used.

$K=C+273 $

If Sirius has a surface temperature of $9667_{∘}C,$ what color does it appear?
b The following literal equation shows the relationship between temperatures in Kelvin $K$ and temperatures in degrees Fahrenheit $F.$

$K=95 (F−32)+273 $

Solve the formula for $F.$
a White

b $F=59 (K−273)+32$

c $5840−8540$

a Substitute the value into the given equation.

b Use the Properties of Equality and inverse operations to isolate $F.$

c Substitute the minimum and maximum range values for a light orange star into the derived formula from Part B.

a To determine the apparent color of Sirius, its surface temperature should be converted into Kelvin. To do so, $9667$ will be substituted into the given formula.

b In the given formula, $F$ is the variable of interest. To isolate $F,$ inverse operations will be used.

$K=95 (F−32)+273$

SubEqn

$LHS−273=RHS−273$

$K−273=95 (F−32)$

MultEqn

$LHS⋅59 =RHS⋅59 $

$59 (K−273)=F−32$

AddEqn

$LHS+32=RHS+32$

$59 (K−273)+32=F$

RearrangeEqn

Rearrange equation

$F=59 (K−273)+32$

c It is known that Pollux appears as a light orange point of light. Therefore, its surface temperature is between $3500$ and $5000$ Kelvin. These values should be substituted into the derived formula from Part B. Start with minimum value $3500.$

$F=59 (K−273)+32$

Substitute

$K=3500$

$F=59 (3500−273)+32$

▼

Evaluate right-hand side

SubTerm

Subtract term

$F=59 (3227)+32$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$F=529043 +32$

CalcQuot

Calculate quotient

$F=5808.6+32$

AddTerms

Add terms

$F=5840.6$

RoundInt

Round to nearest integer

$F≈5841$

$F=59 (K−273)+32$

Substitute

$K=5000$

$F=59 (5000−273)+32$

▼

Evaluate right-hand side

SubTerm

Subtract term

$F=59 (4727)+32$

MoveRightFacToNum

$ca ⋅b=ca⋅b $

$F=542543 +32$

CalcQuot

Calculate quotient

$F=8508.6+32$

AddTerms

Add terms

$F=8540.6$

RoundInt

Round to nearest integer

$F≈8541$

$Range(_{∘}F) 5840−8540 $

Pop Quiz

Use the Properties of Equality to solve the given literal equation for the indicated variable. Some equations may require dividing both sides of the equation by a variable. Since division by zero is not defined, for these equations assume that the variable is not equal to $0.$

Example

On Sunday Dominika spends time on her hobby, which is candle making. She uses $170$ cubic centimeters of wax to produce a cylindrical candle of radius $3$ centimeters.

The volume of a cylinder with radius $r$ and height $h$ is given by the following formula.$V=πr_{2}h $

a Solve the formula for $h.$

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a Use the Division Property of Equality.

b Substitute the given values into the derived formula.

a On the right-hand side of the formula, $h$ is multiplied by $πr_{2}.$

$V=πr_{2}h $

To isolate $h,$ the inverse operation of multiplication should be applied to both sides of the equation. This means applying the Division Property of Equality. $V=πr_{2}h$

DivEqn

$LHS/πr_{2}=RHS/πr_{2}$

$πr_{2}V =πr_{2}πr_{2}h $

CancelCommonFac

Cancel out common factors

$πr_{2}V =πr_{2}πr_{2}h $

SimpQuot

Simplify quotient

$πr_{2}V =h$

RearrangeEqn

Rearrange equation

$h=πr_{2}V $

b It is known that $170$ cubic centimeters of wax were used, and that the radius of the cylindrical candle is $3$ centimeters. Therefore, $V=170$ and $r=3$ will be substituted into the derived formula.

Closure

A literal equation shows the relationship of two or more variables. In some cases, a literal equation needs to be solved for a specific variable. Coming back to the challenge presented at the beginning of the lesson, the time Maya spent jogging can also be found by solving the given formula for $t.$
### Answer

### Hint

### Solution

Now, substitute $d=150$ and $v=2.5.$

$d=v⋅t $

It is known that Maya will run $150$ meters at a speed of $2.5$ meters per second.
a Determine how long it takes Maya to jog this distance by trying some values for some $t-$values until a true statement is obtained.

b Find a different way to solve the question.

a **Table:**

Time (s) | Distance Traveled (m) |
---|---|

$10$ | $25$ |

$20$ | $50$ |

$30$ | $75$ |

$40$ | $100$ |

$50$ | $125$ |

$60$ | $150$ |

It takes Maya $60$ seconds to run $150$ meters.

b See solution.

a Use multiples of $10$ for $t$ to make a table.

b Solve the formula for $t$ and substitute the given values.

a It is known that Maya's speed is $2.5$ meters per second. By the formula, the speed multiplied by the time equals the distance traveled. For different $t-$values, the distance traveled is shown in the table.

Time (s) | Speed $×$ Time | Distance Traveled (m) |
---|---|---|

$10$ | $2.5(10)=25$ | $25$ |

$20$ | $2.5(20)=50$ | $50$ |

$30$ | $2.5(30)=75$ | $75$ |

$40$ | $2.5(40)=100$ | $100$ |

$50$ | $2.5(50)=125$ | $125$ |

$60$ | $2.5(60)=150$ | $150$ |

Maya will run $150$ meters in $60$ seconds at a speed of $2.5$ meters per second.

b The same result can be found by solving the formula for $t$ and then substituting in the given values.

$d=v⋅t$

▼

Solve for $t$

DivEqn

$LHS/v=RHS/v$

$vd =vv⋅t $

CancelCommonFac

Cancel out common factors

$vd =v v ⋅t $

SimpQuot

Simplify quotient

$vd =t$

RearrangeEqn

Rearrange equation

$t=vd $