Question

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

3C+4D=5

2C+5D=2

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?

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.

Answer 1

Answer:

C = 5/3 -4D/3    

C = 1 – 5D/2  

D= -4/7

C=  17/7  

Explanation:

As per question statement we have following equations;

3C+4D=5   ……….. eq 1

2C+5D=2  …………eq2

In eq 1, lower coefficient is with C, so we will isolate it ;

3C+4D=5  

3C=5-4D;

C = (5-4D)/3

C = 5/3 -4D/3    ………..eq3

In eq 2, lower coefficient is with C, so we will isolate it

2C+5D=2

2C = 2 - 5D

C = (2 – 5D)/2;

C = 1 – 5D/2  ………… eq4

Equating/substituting eq3 and eq4;

C = 5/3 -4D/3

C = 1 – 5D/2

 -     -    +

____________      

 0 = (5/3-1) + (-4D/3 + 5D/2)

 0 = (2/3) + (7D/6)

 Multiplying by 6;

 0 = 4 + 7D;

D= -4/7

Putting D in eq 4;

C = 1 – 5D/2

C = 1 – 5/2x(4/7)

C= 1 + 20/14

C = 1+10/7

C=  17/7