if y=x-3, then you have to plug it into the equation.
for numb 1, y=x-3 so when you plug that in it makes the equation x+y=5 turns into x+x-3=5. then you solve by making the sides equal.
you do this by x+x-3=5
x-3=3
x=2
please help, I have a test tomorrow and i don't get this.
solve by substitution
1. y = x - 3
x + y = 5
2. 5x + 2y = 0
x - 3y =0
2 answers
I'm with Zava on the idea, but I think one of us dropped a sign along the way. And I'm not at all sure it wasn't me, so I'll check at the end. Dropped signs happen all the time.
We have:
x + y = 5
but we're also given what y is: y = x - 3
So we can just replace y in the equation with (x-3)
x + y = 5
x + (x-3) = 5
2x - 3 = 5
2x = 5 + 3
2x = 8
x = 4
and since y = x -3
y = 4 - 3 = 1
Check:
y = x - 3
1 = 4 - 3 - OK
x + y = 5
4 + 1 = 5 - OK
The second one is a leetle trickiersince you''ll have to make one of the equations work for the substitution
x - 3y = 0
-> x = 3y --- that's the one we'll use!
5x + 2y = 0
5(3y) + 2y = 0
17y = 0 -> y = 0.
No that's not a mistake. The only answer is if x and y are both zero!
We have:
x + y = 5
but we're also given what y is: y = x - 3
So we can just replace y in the equation with (x-3)
x + y = 5
x + (x-3) = 5
2x - 3 = 5
2x = 5 + 3
2x = 8
x = 4
and since y = x -3
y = 4 - 3 = 1
Check:
y = x - 3
1 = 4 - 3 - OK
x + y = 5
4 + 1 = 5 - OK
The second one is a leetle trickiersince you''ll have to make one of the equations work for the substitution
x - 3y = 0
-> x = 3y --- that's the one we'll use!
5x + 2y = 0
5(3y) + 2y = 0
17y = 0 -> y = 0.
No that's not a mistake. The only answer is if x and y are both zero!
0 answers