How do you solve the system 1/3x-y=313xy=3 and 2x+y=252x+y=25 using substitution?

1 Answer
Oct 2, 2017

x = 12x=12 and y=1y=1

Explanation:

Given equations are:

(1) ------ 1/3x-y=313xy=3

(2)------2x+y=252x+y=25

Multiply (1) by 3 to eliminate the fractional part,

(1) ------- x- 3y = 9x3y=9

=> x= 9+ 3yx=9+3y -------- let this be equation (3)

Now substitute this value of xx from (3) in equation (2),

(2) --------- 2(9+3y) + y = 252(9+3y)+y=25

=> 18 + 6y +y = 2518+6y+y=25

=> 7y = 25- 187y=2518

=> y = 7/7y=77

Therefore,

y=1y=1

Substituting this value of yy in equation (3),

x= 9 + 3\times 1x=9+3×1

x = 9+3 x=9+3

Therefore,
x = 12x=12
So we have the values of x x and yy as,

x = 12x=12 and y=1y=1