Temperature Conversion

Temperature Conversion exercise

• Rewrite the Fahrenheit equation and set it equal to Celsius.

  Hints

  The inverse of addition is subtraction.

  The reciprocal is also called the Multiplicative Inverse. It's the number we can multiply by in order to get $1$. To get a reciprocal of a number, we change the numerator and denominator.

  For example: the reciprocal of $7$ is $\frac{1}{7}$ because $7 \cdot \frac{1}{7}=1$. The reciprocal of $\frac{4}{11}$ is $\frac{11}{4}$ because $\frac{4}{11} \cdot \frac{11}{4} = 1$.

  Only like terms can be simplified when adding and subtracting. If they are not like terms, leave it as an expression.

  Solution

  Nigel has to isolate $C$ and since $32$ is being added, Nigel needs to subtract $32$ from both sides of the equation.

  • $\begin{array}{rcr} F&=&\frac{9}{5}C+32\\ \color{#669900}{-32} & &\color{#669900}{-32}\\\end{array}$
  • Since 32-32=0, they cancel each other out
  • Now the equation is written as $F-32=\frac{9}{5}C$

  Nigel is one step away from isolating $C$ ! To isolate $C$, he multiplies both sides of the equation by the reciprocal of $\frac{9}{5}$, which is $\frac{5}{9}$ .

  • $\begin{array}{rcr} (F-32) \color{#669900}{\cdot\frac{5}{9}}=\frac{9}{5}C \color{#669900}{\cdot\frac{5}{9}}\\ \end{array}$
  • $\frac{5}{9}\cdot \frac{9}{5}=1$ cancel each other out

  The equation is now set equal to C, which is $C=\frac{5}{9}(F-32)$.

• Determine the corresponding unit conversions.

  Hints

  The Fahrenheit temperature is always a larger number than its equivalent Celsius temperature.

  The Fahrenheit and Celsius scales begin at a different number - where $0^{\circ}$ Celsius is freezing, that equivalent temperature in Fahrenheit is $32^{\circ}$.

  Recall the formulas:

  • $F=\frac{9}{5}C+32$
  • $C=\frac{5}{9}(F-32)$

  Solution

  First, decide which formula you are going to use:

  $F=\frac{9}{5}C+32$ or $C=\frac{5}{9}(F-32)$

  Pick one of the formulas - you do not have to use both. If you really want to use both, you can use the second formula to check your answer!

  To convert the temperature of the living room from Fahrenheit to Celsius, use the formula $C=\frac{5}{9}(F-32)$ and plug in $72$ for F.

  • $C=\frac{5}{9}(72-32)=\frac{5}{9}(40)=22.2^{\circ}$

  • If you want to use the other formula, you will see that the pairs still match:

  • $F=\frac{9}{5}(22.2)+32=39.96+32=72^{\circ}$

  To convert the temperature of the kitchen from Fahrenheit to Celsius, use the formula $C=\frac{5}{9}(F-32)$ and plug in $68^{\circ}$ for F.

  • $C=\frac{5}{9}(68-32)=\frac{5}{9}(36)=20^{\circ}$

  To convert the temperature of the shower from Fahrenheit to Celsius, use the formula $C=\frac{5}{9}(F-32)$ and plug in $110^{\circ}$ for F.

  • $C=\frac{5}{9}(110-32)=\frac{5}{9}(78)=43^{\circ}$

  To convert the temperature of the freezer from Fahrenheit to Celsius, use the formula $C=\frac{5}{9}(F-32)$ and plug in $0^{\circ}$ for F.

  • $C=\frac{5}{9}(0-32)=\frac{5}{9}(-32)= -17.7^{\circ}$
  • Yes, temperatures can be negative numbers!

  To convert the temperature of the bedroom from Fahrenheit to Celsius, use the formula $C=\frac{5}{9}(F-32)$ and plug in $75^{\circ}$ for F.

  • $C=\frac{5}{9}(75-32)=\frac{5}{9}(43)=23.9^{\circ}$

• Estimate the conversion between Fahrenheit and Celsius.

  Hints

  Since $\frac{9}{5}\approx2$, you can replace $\frac{9}{5}$ with $2$ when you are estimating the conversion from Celsius to Fahrenheit.

  Since $\frac{5}{9}\approx\frac{1}{2}$, you can replace $\frac{5}{9}$ with $\frac{1}{2}$ when you are estimating the conversion from Fahrenheit to Celsius.

  When converting Celsius to Fahrenheit, you will have to multiply by a value greater than $1$ and add a positive value. This means that the converted Fahrenheit value will always be greater than the Celsius value.

  Solution

  For the shower: To estimate the conversion from Celsius to Fahrenheit, Nigel has to multiply $43$ degrees Celsius by $2$, which equals $86$. Then he has to add a value of $32$, which then equals $118°F$.

  For the tea: To estimate the conversion from Fahrenheit to Celsius, Nigel will subtract $32$ from $170$, which equals $138$. Then he divides that value by $2$, which results to $69°C$.

• Rewrite the equation.

  Hints

  To eliminate a term that is being added, take the inverse operation which is subtraction.

  Dividing by a value is the same as multiplying by the reciprocal of that value. For example: multiplying by sides by $\frac{1}{2}$ is the same as dividing by $2$.

  If terms do not combine, then write them as an expression. For example, $x+10$ would stay as $x+10$.

  Solution

  Step $1$: Since Nigel knows that the width of the room is $10$ ft, we first plug in $10$ for $W$.

  • $P=2(L+W)$
  • $P=2(L+10)$

  Step $2$: Divide both sides of the equation by $2$. The $2$ is canceled on the right side.

  • $\begin{array}{rcl}P&=&2(L+10)\\\color{#669900}{\div2} & & \color{#669900}{\div2}\\\end{array}$
  • $\frac{P}{2}= (L+10)$

  Step $3$: Subtract $10$ on both sides. The $10$ is canceled on the right side.

  • $\begin{array}{rcr} \frac{P}{2}&=& L+10\\ \color{#669900}{-10} & &\color{#669900}{-10}\\ \end{array}$
  • $\frac{P}{2}-10=L$

  Step $4$: Use the symmetric property to move $L$ to the left side of the equation. Remember, the symmetric property of equality says that if $a = b$, then $b = a$.

  • $L=\frac{P}{2}-10$

• Given the value of one variable, solve the equation.

  Hints

  Plug in the given value for $b$, then follow order of operations to simplify that side.

  Then use the inverse operations to isolate a variable. For example, the inverse of addition is subtraction, and the inverse of multiplication is division.

  To solve for $a$, the variable has to be completely isolated with a coefficient of positive $1$.

  Solution

  • First, plug in the given value for $b$ into the given equation.
  • Next, follow order of operations: parentheses, multiplication/division, and addition/subtraction. Keep in mind that when you are multiplying/dividing or adding/subtracting, you should work from left to right.
  • Simplify until the equation is solved for $a$.

  In order from lowest value to greatest value:

  Problem 1

  • $2a = 6(b-19)$ where $b= 5$
  • $2a = 6(5-19)$
  • $2a = 6(-14)$
  • $2a= -84$
  • Since the coefficient of $a$ is not yet one, we need to divide both sides by $2$, leaving us with:
  • $a = -42$

  Problem 2

  • $\frac{3}{8}b-23=a$ where $b=11$
  • $a=\frac{3}{8}(11)-23$
  • $a=\frac{33}{8}-23$
  • $a=\frac{33}{8}-\frac{184}{8}$
  • $a=-\frac{151}{8}$

  Problem 3

  • $a=\frac{5}{4}(b-7)$ where $b= 0$
  • $a=\frac{5}{4}(0-7)$
  • $a=\frac{5}{4}(-7)$
  • $a=-\frac{35}{4}$

  Problem 4

  • $a=12b-40$ where $b=6$
  • $a=12(6)-40$
  • $a= 72-40$
  • $a= 32$

  Problem 5

  • $a+7b=18$ where $b= -9$
  • $a+7(-9)=18$
  • $a+(-63)=18$
  • $a=81$

• Match the equivalent two-variable equations.

  Hints

  When there is an expression being divided by a value, you can multiply the entire expression by that value to eliminate the denominator.

  For example:

  • $y=\frac{a+6}{3}$
  • $3\cdot y=3\cdot \frac{a+6}{3}$
  • $3y= a+6$
  • $3y-6 = a$

  When you have a constant multiplied to an expression inside of parentheses, you can eliminate that constant by dividing both sides of the equation by that constant.

  For example:

  • $y= 5(x+7)$
  • $\frac{y}{5} = \frac{5(x+7)}{5}$
  • $\frac{y}{5}= x+7$
  • $\frac{y}{5} -7 = x$

  In order to completely isolate a variable, the coefficient of the variable must be positive $1$. If it is a negative value, then you have to multiply or divide by $-1$ on both sides of the equation.

  For example:

  • $7y= -x$
  • $-1\cdot 7y=-1\cdot (-x)$
  • $-7y= x$

  Solution

  Starting with $x=3y+7$ and $y=\frac{x-7}{3}$ , we want to isolate $y$, so we need to use reverse order of operations and subtract $7$ from both sides of the equation and simplify.

  • $\begin{array}{rcr} x&=&3y+7\\ \color{#669900}{-7} & &\color{#669900}{-7}\\ \end{array}$
  • $x-7=3y$

  Next, we divide both sides of the equation by $3$ and simplify.

  • $\begin{array}{rcr} x-7&=&3y\\ \color{#669900}{\div3} & & \color{#669900}{\div3}\\ \end{array}$
  • $\frac{x-7}{3}=y$

  Notice that the $y$ is now isolated. We could leave it like this or we could use the symmetric property of equality and write:

  • $y=\frac{x-7}{3}$

  ----------------------------------------------------------------------------------

  Starting with $x=\frac{18}{4}y-4$ , we first have to add $4$ to both sides to isolate y.

  • $\begin{array}{rcr} x&=&\frac{18}{4}y-4\\ \color{#669900}{+4} & &\color{#669900}{+4}\\ \end{array}$
  • $x+4=\frac{18}{4}y$

  Multiply both sides by the reciprocal of $\frac{18}{4}$ , which is $\frac{4}{18}$ .

  • $\begin{array}{rcr} (x+4)\color{#669900}{\cdot\frac{4}{18}}=\frac{18}{4}y \color{#669900}{\cdot\frac{4}{18}}\\ \end{array}$
  • $\frac{4}{18}(x+4)= y$

  We can take this a step further and reduce the fraction $\frac{4}{18}$ .

  • $\frac{4}{18} = \frac{2}{9}$

  Notice that the $y$ is now isolated. We could leave it like this ,or we could use the symmetric property of equality and write: $y= \frac{2}{9}(x+4)$

  -------------------------------------------------------------------------------------

  To isolate y for the equation $x=3(y+4)$ ,we first have to divide both sides of the equation by $3$ .

  • $\begin{array}{rcr }x&=&3(y+4)\\ \color{#669900}{\div3} & & \color{#669900}{\div3}\\ \end{array}$
  • $\frac{x}{3}=y+4$

  The last step is to subtract $4$ on both sides.

  • $\begin{array}{rcr}\frac{x}{3} &=& y+4\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ \end{array}$
  • $y=\frac{x}{3}-4$

  --------------------------------------------------------------------------

  To isolate y for the equation $x=20-5y$ , we first have to subtract both sides of the equation by $20$ .

  • $\begin{array}{rcl} x&=&20-5y\\ \color{#669900}{-20} & &\color{#669900}{-20}\\ \end{array}$
  • $x-20= -5y$

  Next, divide both sides of the equation by $-5$ .

  • $\begin{array}{rcr} x-20&=&-5y\\ \color{#669900}{\div(-5)} & &\color{#669900}{\div(-5)}\\ \end{array}$
  • $\frac{x-20}{-5}=y$

  Notice that the $y$ is now isolated. We can take this a step further by separating the left side into two fractions and simplify:

  • $-\frac{x}{5} + \frac{20}{5}$ = $-\frac{x}{5} + 4$

  We could leave it like this or we could use the symmetric property of equality and write:

  • $y=-\frac{x}{5} + 4$

  ------------------------------------------------------------------------------------

  To isolate y for the equation $x=\frac{3}{7}(y+3)$ ,we first have to multiply both sides of the equation by the reciprocal of $\frac{3}{7}$ which is $\frac{7}{3}$ .

  • $\begin{array}{rcr} x\color{#669900}{\cdot\frac{7}{3}}=\frac{3}{7}(y+3) \color{#669900}{\cdot\frac{7}{3}}\\ \end{array}$
  • $\frac{7}{3}x=y+3$

  Next, subtract both sides of the equation by $3$ .

  • $\begin{array}{rcr}\frac{7}{3}x&=&y+3\\ \color{#669900}{-3} & &\color{#669900}{-3}\\ \end{array}$
  • $ \frac{7}{3}x-3=y$

  Notice that the $y$ is now isolated. We could leave it like this or we could use the symmetric property of equality and write:

  • $y=\frac{7}{3}x-3$