Question

Part D- Isolating a Variable with a Coefficient In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have • 30 + 4D = 5 • 2C + 5D = 2 t to None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?

1. D= 5/4 - 3/4C
2. C= 5/3 - 4/3D
Part E- Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers. Enter the answer as two numbers (either fraction or decimal), separated by a comma, with C first.

Answer 1

D) D =
`(5)/(4) - (3)/(4) \ C`, E) (C, D) = (
`(17)/(7), \ (-4)/(7)`

Step-by-step explanation:

Part D) two expressions are indicated

3C + 4D = 5

2C +5 D = 2

let's simplify each expression

3C + 4D = 5

4D = 5 - 3C

we divide by 4

D =
`(5)/(4) - (3)/(4) \ C`

The other expression

2C +5 D = 2

2C = 2 - 5D

C =
`1 - (5)/(2) \ D`

we can see that the correct result is 1

Part E.

It is asked to solve the problem by the substitution method, we already have

D =
`(5)/(4) - (3)/(4) \ C`

we substitute in the other equation

2C +5 D = 2

2C +5 (5/4 - ¾ C) = 2

we solve

C (2 - 15/4) + 25/4 = 2

-7 / 4 C = 2 - 25/4

-7 / 4 C = -17/4

7C = 17

C =
`(17)/(7)`

now we calculate D

D =
`(5)/(4) - (3)/(4) \ (17)/(7)`

D = 5/4 - 51/28

D =
`(35-51)/(28)`

D = - 16/28

D =
`- (4)/(7)`

the result is (C, D) = (
`(17)/(7), \ (-4)/(7)` )