Algebra 1 > Notes > Yorkcounty Final > Unit 3 > Lesson_8 - Solving Literal Equations

Lesson 8 Solving Literal Equations

Quick Reminder on Solving  

Remember that when you are solving for one variable you want to get that thing by itself on one side of the equation. What you do to one side of the equation you must do to the other side.

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When solving for that one variable add and subtract things you want to move first. Do not try to multiply or divide until you have moved everything you can with addition and subtraction.

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Then multiply or divide the rest to get your variable alone.

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If you are trying to solve for a; 2a + 5 = 10, you want to move the ( + 5 ) by doing the opposite, which is, subtracting 5 ( - 5 ) to both sides. 2a + 5 – 5 = 10 – 5. Then you will have 2a = 5. Now the 2 being multiplied to your a is keeping a from being alone. Now you want to divide by sides by 2 to get a alone. This produces a = 5/2

Literal Equations  Literal

Equations are equations with more than one variable in them.  In this lesson you are going to solve for one of the variable.  This means you are going to try to get one variable by itself on one side of the equation while moving everything else to the other side.

Example 1 

a + b = c ; solve for a

 We

need to get a by itself on the left side so we need to move b over. (subtract b from both sides)  a+b–b=c–b  a=c–b (since c and b are different we cannot subtract them, so we just write that as “c – b”)  So, a = c - b

Try this 

a + 3b – d = 4; solve for b

 Solution: 

(remember, get b alone)

(move stuff with addition/subtraction first then multiply/divide last)

a

– a + 3b – d = 4 – a  3b – d = 4 – a  3b – d + d = 4 – a + d

(subtract a from both sides)

(add d to both sides) (since 3, -a and d are different we cannot

combine them)

 3b

=4–a+d

(what is keeping b from being alone? The 3

being multiplied to it, so we must divide by 3)

3b 4 − a + d = 3 3

so .

b=

4−a+d 3

Example 2  Last

lesson we talked about the perimeter of a rectangle. The formula for that was P= 2l + 2w

 Using  Try

P = 2l + 2w, solve for w

this problem now, continue to the next slide when you are done.

Example 2 P

= 2l + 2w

P

(we want to solve for w, so we need to get w alone)

– 2l = 2l – 2l + 2w  P – 2l = 2w

P − 2l 2 w = 2 2  So

P − 2l w= 2

(first subtract 2l from both sides) (now we need to divide by 2 to get w alone)

Example 3 Solve for a Try this one on your own, next slide for answer.

1 a + b = 2c 2

Example 3 1 a + b = 2c 2

First, subtract b from both sides

1 a + b − b = 2c − b 2

1 a = 2c − b 2

Second, multiply both sides by 2

1 2 a = 2( 2c − b ) 2 a = 4c – 2b

Remember, when you multiply both sides both 2, everything on both sides gets multiplied by 2. So 2 multiplies 2c-b

Concepts to remember…  When

you are solving for one variable move things with addition and subtraction first.  When you have moved everything you can with addition and subtraction then move the rest of the things with multiplication and division. DO NOT USE MULTIPLICATION OR DIVISION until you have moved everything you can with addition and subtraction.