Urgent Math Help- Solving using Elimination?

Multiply the first equation by 4.

16a/3 - b = 24

5a/6 + b = 13

Now you can get rid of the b's since one is negative and one is positive. Add the a's and the c's.

16a/3 + 5a/6 = 37

32a/6 + 5a/6 = 37

37a/6 = 37

37a = 222

a = 6

Now plug in 6 for a in either equation. The second one is easier.

5(6)/6 + b = 13

5 + b = 13

b = 8

4a/3 - b/4 = 6 --- eq1

5a/6 + b = 13 --- eq2

to solve systems using elimination, one variable must have the same coefficient with same signs (if elimination by subtraction) or same coefficient with different signs (if by addition)

since your given has rational/fraction coefficients, find the LCD to eliminate denominators (but not ruining the real equation). do it with eq1 then eq2

4a/3 - b/4 = 6 ... LCD = 3 * 4 = 12. multiply each term by 12

12(4a/3) - 12(b/4) = 12(6)

16a - 3b = 72 ----> eq3

5a/6 + b = 13 ... LCD = 6 . multiply each term by 6

6(5a/6) + 6(b) = 13(6)

5a + 6b = 78 -----> eq4

now make one variable have same coefficients. multiply eq3 by 2.

(16a - 3b = 72)2

32a - 6b = 144 -----> eq5

now add eq4 and eq5 to eliminate b and solve for a

5a + 6b = 78

32a - 6b = 144

37a = 222

a = 6

now to solve for b, just substitute b from either eq3, eq4, etc.

good luck!

16a - 3b = 72

5a + 6b = 78

32a - 6b = 144

5a + 6b = 78______ADD

37a = 222

a = 6

30 + 6b = 78

6b = 48

b = 8

a = 6 , b = 8