Question

Solve the following system of equations algebraically:

30x + 44y = 10; 40x + 55y = 13

Answer 1

30x + 44y = 10   ...(1)

40x + 55y = 13   ...(2)

From equation (1)

30x = 10 − 44у

x = `(10 - 44y)/30`   ...(3)

Substitute the value of x in equation (2).

`40((10 - 44y)/30) + 55y = 13`

`(40 - 176y)/3 + (55y)/1 = 13`

`(40 - 176 + 165y)/3 = 13`

40 − 11у = 39

40 − 39 = 11y

y = `1/11`

From equation (3)

x = `(10 - 44(1/11))/30`

= `(10 - 4)/30`

= `6/30`

= `1/5`

∴ x = `1/5`, y = `1/11`