4 literal equations

Literal Equations

Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation.

Literal Equations

Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation.

Literal Equations

Example A.

a. Solve for x if x + b = c

b. Solve for w if yw = 5.

Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations.

Example A.

a. Solve for x if x + b = c

Literal Equations



Example A.

a. Solve for x if x + b = c Remove b from the LHS by subtracting from both sides x + b = c –b –b x = c – b

Literal Equations


Example A.


Literal Equations

b. Solve for w if yw = 5. Remove y from the LHS by dividing both sides by y.


Example A.


Literal Equations

b. Solve for w if yw = 5. Remove y from the LHS by dividing both sides by y. yw = 5 yw/y = 5/y


Example A.


5y

Literal Equations

b. Solve for w if yw = 5. Remove y from the LHS by dividing both sides by y. yw = 5 yw/y = 5/y w =


Literal EquationsAdding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.

To solve for a specific variable in a simple literal equation, do the following steps.


To solve for a specific variable in a simple literal equation, do the following steps.1. If there are fractions in the equations, multiple by the LCD to clear the fractions.


To solve for a specific variable in a simple literal equation, do the following steps.1. If there are fractions in the equations, multiple by the LCD to clear the fractions.2. Isolate the term containing the variable we wanted to solve for


To solve for a specific variable in a simple literal equation, do the following steps.1. If there are fractions in the equations, multiple by the LCD to clear the fractions.2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation.


To solve for a specific variable in a simple literal equation, do the following steps.1. If there are fractions in the equations, multiple by the LCD to clear the fractions.2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. 3. Isolate the specific variable by dividing the rest of the factor to the other side.



Literal Equations

Example B. a. Solve for x if (a + b)x = c

Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.


Literal Equations

Example B. a. Solve for x if (a + b)x = c (a + b) x = c

div the RHS by (a + b)



Literal Equations

Example B. a. Solve for x if (a + b)x = c (a + b) x = c x = c

(a + b)

div the RHS by (a + b)


b. Solve for w if 3y2w = t – 3Literal Equations

b. Solve for w if 3y2w = t – 3

3y2w = t – 3 div the RHS by 3y2

Literal Equations


3y2w = t – 3 w = t – 3

3y2

div the RHS by 3y2

Literal Equations


3y2w = t – 3

c. Solve for a if b2 – 4ac = 5

w = t – 3 3y2

div the RHS by 3y2

Literal Equations


3y2w = t – 3


b2 – 4ac = 5

w = t – 3 3y2

div the RHS by 3y2

move b2 to the RHS

Literal Equations


3y2w = t – 3


b2 – 4ac = 5 – 4ca = 5 – b2

w = t – 3 3y2

div the RHS by 3y2

move b2 to the RHS

Literal Equations


3y2w = t – 3


b2 – 4ac = 5 – 4ca = 5 – b2

w = t – 3 3y2

div the RHS by 3y2

move b2 to the RHS div the RHS by –4c

Literal Equations


3y2w = t – 3


b2 – 4ac = 5 – 4ca = 5 – b2

w = t – 3 3y2

a = 5 – b2

–4c

div the RHS by 3y2


Literal Equations


3y2w = t – 3


b2 – 4ac = 5 – 4ca = 5 – b2

w = t – 3 3y2

a = 5 – b2

–4c

div the RHS by 3y2


Literal Equations

d. Solve for y if a(x – y) = 10


3y2w = t – 3


b2 – 4ac = 5 – 4ca = 5 – b2

w = t – 3 3y2

a = 5 – b2

–4c

div the RHS by 3y2


Literal Equations


a(x – y) = 10 expand


3y2w = t – 3


b2 – 4ac = 5 – 4ca = 5 – b2

w = t – 3 3y2

a = 5 – b2

–4c

div the RHS by 3y2


Literal Equations



ax – ay = 10


3y2w = t – 3


b2 – 4ac = 5 – 4ca = 5 – b2

w = t – 3 3y2

a = 5 – b2

–4c

div the RHS by 3y2


Literal Equations



ax – ay = 10 subtract ax


3y2w = t – 3


b2 – 4ac = 5 – 4ca = 5 – b2

w = t – 3 3y2

a = 5 – b2

–4c

div the RHS by 3y2


Literal Equations




– ay = 10 – ax


3y2w = t – 3


b2 – 4ac = 5 – 4ca = 5 – b2

w = t – 3 3y2

a = 5 – b2

–4c

div the RHS by 3y2


Literal Equations




– ay = 10 – ax div by –a


3y2w = t – 3


b2 – 4ac = 5 – 4ca = 5 – b2

w = t – 3 3y2

a = 5 – b2

–4c

div the RHS by 3y2


Literal Equations




– ay = 10 – ax div by –a

y = 10 – ax –a

Multiply by the LCD to get rid of the denominator then solve.

Literal Equations

–4 = d3d + b


Example C. Solve for d if

Literal Equations

–4 = d3d + b


multiply by the LCD d– 4 = d3d + b


Literal Equations

–4 = d3d + b


multiply by the LCD d

d

– 4 = d3d + b

– 4 = d3d + b( )

d


Literal Equations

–4 = d3d + b



d

– 4 = d3d + b

– 4 = d3d + b( )

d


Literal Equations

–4 = d3d + b



d

– 4 = d3d + b

– 4 = d3d + b( )

d

– 4d = 3d + b


Literal Equations

–4 = d3d + b



d

– 4 = d3d + b

– 4 = d3d + b( )

d

– 4d = 3d + b move –4d and b


Literal Equations

–4 = d3d + b



d

– 4 = d3d + b

– 4 = d3d + b( )

d

– 4d = 3d + b

– b = 3d + 4d

move –4d and b


Literal Equations

–4 = d3d + b



d

– 4 = d3d + b

– 4 = d3d + b( )

d

– 4d = 3d + b

– b = 3d + 4d

move –4d and b

– b = 7d


Literal Equations

–4 = d3d + b



d

– 4 = d3d + b

– 4 = d3d + b( )

d

– 4d = 3d + b

– b = 3d + 4d

move –4d and b

– b = 7d

= d 7–b


Literal Equations

div. by 7

Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation.

Literal Equations

x = 3 + 12y

Example D. Solve for y if

Flipping EquationsLiteral Equations

Flipping EquationsLiteral Equations

x = 3 + 12y


Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation.

Literal Equations

x = 3 + 12y


= DC

Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular:

Literal Equations

BA

= CD

AB

Reciprocateboth sides

x = 3 + 12y


= DC


x = 3 + 12y


Literal Equations

BA

= CD

AB

x = 3 + 12y


= DC


x = 3 + 12y


Literal Equations

BA

= CD

AB

x = 3 + 12y

x – 3 = 12y


= DC


x = 3 + 12y


Literal Equations

BA

= CD

AB

x = 3 + 12y

x – 3 = 12y reciprocating both sides,

= 12y

x – 31


= DC


x = 3 + 12y


Literal Equations

BA

= CD

AB

x = 3 + 12y


then div. by 2= 12y

x – 31

= 2y2(x – 3)

1


2

= DC


x = 3 + 12y


Literal Equations

BA

= CD

AB

x = 3 + 12y


then div. by 2= 12y

x – 31

= 2y2(x – 3)

1


2 = y or 2(x – 3)1y =

= DC


x = 3 + 12y


Literal Equations

BA

= CD

AB

x = 3 + 12y


then div. by 2= 12y

x – 31

= 2y2(x – 3)

1


2 = y or 2(x – 3)1y =

Be careful with the flip,

1y

+1x

11y

+1x

reciprocates

(which is not x + y).

Exercise. Solve for the indicated variables.

Literal Equations

1. a – b = d – e for b. 2. a – b = d – e for e.

3. 2*b + d = e for b. 4. a*b + d = e for b.

5. (2 + a)*b + d = e for b. 6. 2L + 2W = P for W

7. (3x + 6)y = 5 for y 8. 3x + 6y = 5 for y

w = t – 3 613. for t w = t – b

a14. for t

w =11. for t w =12. for tt6

6t

w = 3t – b a15. for t 16. (3x + 6)y = 5 for x

w = t – 3 617. + a for t w = at – b

518. for t

w = at – b c19. + d for t 3 = 4t – b

t20. for t

9. 3x + 6xy = 5 for y 10. 3x – (x + 6)y = 5z for y

Education

4 literal equations