The sum of 4 times a larger integer and 5 times a smaller integer is 7. When twice the smaller integer is subtracted from 3 times the larger, the result is 11. Find the integers.

Answer 1

x is the larger, y is the smaller

4x + 5y = 7

3x - 2y=11

2*eq1 + 5 * eq2:

8x+10y +5(3x-2y) = 2*7+5*11

8x + 10y +15x -10y = 14 + 55

23x = 69

x=3

y=-1

the larger is 3, the smaller is -1

Answer 2

Let x = larger integer and y = smaller integer

4x + 5y = 7

3x - 2y = 11

Multiply the first equation by 2 and the second by 5:

8x + 10y = 14

15x - 10y = 55

Add the equations to get 23x = 69

x = 3 and y = -1

Answer 3

4L+5S = 7

3L-2S = 11

multiply 1st by 3 and 2nd by 4 to get

12L-8S = 44

12L+15S=21 subtract to get

-23S = 23

S =-1

3L=11+2S = 9

L = 3

integers are -1 and 3