first one times 3
second one times 5
15 x + 9 y = 12
15 x + 20 y = 45
------------------subtract
0 x - 11 y = - 33
y= 3
then go back for x
5 x + 9 = 4
5 x = -5
x = -1
Solve
5x + 3y = 4
3x + 4y = 9
Just a bit confused
Please help
2 answers
5x+3y=4 [eqn.1]
3x+4y=9[eqn.2]
now,
x=4-3y/5[eqn.3]
Now putting the value of x in eqn.2
3(4-3y/5)+4y=9
12-9y/5+4y=9
12-9y+20y/5=9
12+11y=45
11y=33
y=3
And x=4-3y/5
x=4-3*3/5
x=4-9/5
x=-1
3x+4y=9[eqn.2]
now,
x=4-3y/5[eqn.3]
Now putting the value of x in eqn.2
3(4-3y/5)+4y=9
12-9y/5+4y=9
12-9y+20y/5=9
12+11y=45
11y=33
y=3
And x=4-3y/5
x=4-3*3/5
x=4-9/5
x=-1