What is a Literal equation?
Literal equations are equations with two or more variables (letters or alphabets) in which each variable can be expressed in terms of the other variables. When solving literal equations, the goal is to isolate one variable and express the solution in terms of the other variables explicitly. In a literal equation, each variable represents a quantity.
We are sometimes given equations in the form of geometric figure formulas, such as P = 4s, where P is the square’s perimeter and S is its side length. P and S are two variables that are expressed in terms of S and this is an example of a literal equation. We cannot get the exact numerical value of a variable in a literal equation.
Why do we need to rewrite?
We isolate the desired variable when we rewrite the literal equation, and we get the final solution when we substitute known values for the variables. This will make our calculations easier.
How do we rewrite a equation?
Step 1: Find the variable in the equation that you need to solve for.
Step 2: Determine the operations that are performed on this variable and the order in which they are performed.
Step 3: To undo operations and isolate the variable, use inverse operations.
Example:
Solve the equation 3y + 7x = 9 for y .
Solution:
Write the equation
3y + 7x = 9
Undo the addition.
Subtraction Property of Equality
3y + 7x – 7x = 9 – 7x
Simplify
3y = 9 – 7x
Undo the multiplication.
Division Property of Equality
\(\frac{3y}{3}~=~\frac{9-7x}{3}\)
Simplify
\(y~=~3-~\frac{7}{3}x\)
What is Temperature conversion?
The process of changing the values of temperature from one unit to another is known as temperature conversion. According to the Kelvin scale, the freezing point of water is 273.15K and the boiling point of water is 373.15K. According to the Fahrenheit scale, the freezing point of water is 32℉ and the boiling point of water is 212℉. According to the Celsius scale, the freezing point of water is 0°C and the boiling point is 100°C.
Conversion of Temperature Between Fahrenheit and Celsius
Fahrenheit to Celsius and Celsius to Fahrenheit formulas are listed below. These formulas will help convert the temperature into required units.
- The temperature conversion formula from Fahrenheit to Celsius is
\(C~=~\frac{5}{9}~(F~-~32)\)
- The Temperature Conversion Formula from Celsius to Fahrenheit is
\(F~=~\frac{5}{9}C+32\)
Conversion of Temperature between Fahrenheit and Kelvin
- The temperature conversion formula from Fahrenheit to Kelvin is
\(K~=~\frac{5}{9}~(F-32)+ 273.15\)
- The temperature conversion formula from Kelvin to Fahrenheit is
\(F~=~\frac{9}{5}~(K-273.15)+ 32\)
Let’s discuss how Temperatures in Fahrenheit F is converted to Kelvin K using the formula
\(K~=~\frac{5}{9}~(F-32)+ 273.15\)
Solve the temperature formula in F.
Solution:
\(K~=~\frac{5}{9}(F-32)+273.15\) Write the temperature formula.
\(K~-~273.15~=~\frac{5}{9}(F-32)+273.15-273.15\) Subtraction property of Equality.
\(K- 273.15~= ~\frac{5}{9}~(F-32)\) Simplify
\(\frac{9}{5}\times(K-273.15)~=~\frac{9}{5}~\times~\frac{5}{9}~(F-32)\) Multiplication property of Equality.
\(\frac{9}{5}\times(K-273.15)~=~(F-32)\) Simplify
\(\frac{9}{5}\times(K-273.15)+32~=~F-32+32\) Additive property of Equality
\(\frac{9}{5}~(K-273.15)+32~=~F\) Simplify
Therefore, the temperature conversion formula from Kelvin to Fahrenheit can be obtained from the Fahrenheit to Kelvin formula. \(F~=~\frac{9}{5}~(K-273.15)+32\)
Solved Examples
Example 1:
Solve the equation \( 5y+4x~=~5\) for \( y\).
Solution:
\( 5y+4x~=~5\) Write the equation.
Undo the addition.
\( 5y+4x-4x~=~5-4x\) Subtraction Property of Equality
\( 5y=~5-4x\) Simplify
Undo the multiplication.
\( \frac{5y}{y}~=~\frac{5-4x}{5}\) Division Property of Equality
\( y~=~1-~\frac{4}{5}~x\) Simplify.
Example 2:
The perimeter of a rectangle is calculated using the formula below.
\( P~=~2l+2w\)
To solve for \( l\), rewrite the formula.
Solution:
\( P~=~2l+2w\) Write the formula.
\( P-2w~=~2l+2w-2w\) Subtraction property of Equality.
\( P-2w~=~2l\) Simplify.
\( \frac{P-2w}{2}~=~\frac{2l}{2}\) Division property of equality.
\( l~=\frac{P-2w}{2}\) Simplify.
Example 3:
The surface area of a cylinder is calculated using the formula below.
\( A~=2\pi~rh~+~2\pi~r^2\)
To solve for \( h\), rewrite the formula.
Solution:
\( A~=2\pi~rh~+~2\pi~r^2\) Write the formula.
\( A-~2\pi~r^2~=~2\pi~rh+2\pi~r^2-2\pi~r^2\) Subtraction property of Equality.
\( A-~2\pi~r^2~=~2\pi~rh\) Simplify
\(\frac{A-~2\pi~r^2}{2\pi~r}~=~\frac{2\pi~rh}{2\pi~r} \) Division property of equality.
\(h~=~\frac{A-~2\pi~r^2}{2\pi~r} \) Simplify.
Example 4:
Which thermometer is registering higher temperatures ?
Solution:
You are given the temperature in degrees Fahrenheit in the first thermometer and the temperature in degrees Celsius in the second thermometer. You are asked to find out which temperature is greater.
Convert the temperature in Celsius to Fahrenheit. Then compare the temperatures.
\(F~=~\frac{5}{9}~C+32 \) Write the formula.
\(=~\frac{5}{9}~(90)+32 \) Substitute 90 for C
\(=~50+32 \) Simplify
\(F~=~82 \) Add
Because 112℉＞82℉, the first thermometer is registering greater temperature.
Celsius to Kelvin conversion: Kelvin = Celsius +273.15. The value 273 is used instead of 273.15.
Linear equations have variables of degree 1 and can have only one variable. Literal equations, on the other hand, have at least two variables.