Math: Systems by Elimination

Question

The sum of 5 times the width of a rectangle and twice its length is 26 units. The difference of 15 times the width and three times the length is 6 units. Write and solve a system of equations to find the length and width of the rectangle.

Answer 1

Let W = width, L=length Answer W=2, L=8

equation one: 5W+2L=26

equation two: 15W-3L=6

rewrite equation one, solving for L to get

L=(26-5W)/2

Substitute that into the 2nd equation to get

15W-3(26-5W)/2=6

Multiply by 2, then Expand and solve for W

30W-3(26-5W)=12

30W-78+15W=12

45W=12+78=90

W=90/45=2

L=(26-5W)/2 = (26-10)/2 =8

Width=2 Length=8

Answer 2

First, write the equations down:

5W + 2L = 26

15W - 3L = 6

Now we need to eliminate a variable. I choose W, because 5 goes into 15 easily. Multiply both sides of the first equation by 3. Now we have:

15W + 6L = 78

15W - 3L = 6

If we subtract the two equations, the W's will go away:

0W + 9L = 72

9L = 72

L = 8

Now we can plug L=24 back into one of our equations to solve for W:

5W + 2L = 26

5W + 2*8 = 26

5W = 10

W = 2