Question

Help please, Don't give me the answer just help me get started.
Solve using the elimination method. 1/3x+1/5y =7, 1/6x–2/5y =-4

Answer 1

You want to eliminate one of the variables by adding or subtracting the 2 equations. So you have to compare the coefficients and find a way to get one of the coefficients the same in both equations. Let's choose x (just random you could have chosen y).

In the first equation, x has a coefficient of 1/3. In the second equation, x has a coefficient of 1/6. Can you multiply one of the equations by a factor so that both have the same coefficient for x?

Answer 2

OK I'll give you more hints. If you multiply the second equation by 2 then you get what?

Answer 3

So I'd multiply both x factors by say 2, that would give me 2(1/3), 2(1/6) 2/6 and 2/12, but if I reduce it that brings me back to the starting 1/3 & 1/6.

Answer 4

nonono only multiply the second equation by 2. so:

2* (1/6x–2/5y =-4)

Answer 5

2*(1/6x-2/5y=-4) = 12x - 12/10 reduced to 1.2y;
12x - 1.2y = -8
am I on the right track?

Answer 6

That's not how you multiply fractions :/ Ignore the denominator. When you multiply a fraction by a number, simply multiply the numerator (top) by the number and leave the denominator (bottom) the same. So for example here you're doing 2*(1/6)x only multiply 2 and 1. Then you get (2/6)x

Answer 7

Man it's bee sooo long doing this stuff, but I am not giving up.
Ok so 2/6 would reduce to 1/3 which would eliminate the "1/3x" in the first set right?

Then back to the second equation 2*(1/6x - 2/5y = -4) would be 2/6x - 4/5y = -8 ?

Answer 8

Yayyy good job! Exactly. So you get

2*(1/6x - 2/5y = -4)
=2/6x - 4/5y = -8 ?
=1/3x - 4/5y = -8

Now you have the same coefficient of x for both equations.

Now do equation 1 MINUS (the new) equation 2. If you do it right, you'll be left with no x

Answer 9

aka
1/3x + 1/5y = 7
- (1/3x - 4/5y = -8)

Answer 10

1/3x + 1/5y = 7
1/3x - 4/5y = -8
1y = -1

Answer 11

Very close! :D 7 - -8 is not -1

Answer 12

Doh! should be 15

Answer 13

Yup! so y=15. Now can you go back and solve for x?

Answer 14

Think so , I just do the same except clear the y first then solve from there. Right?

Answer 15

no now it's easy peasy! Just plug in the value of y into either equation and solve for x

Answer 16

1/3x+1/5y =7
y=15
(1/3)x + (1/5)(15) = 7

Answer 17

so (1/3)x + (1/5)(15) = 7
15/3x + 15/5 = 7 ; 5 + 3 = 7
(1/6)x - (2/5) (15) = 15/6 or 2.5 - 30/5 or 6
2.5x - 6 = -4
x = -3.5
so the answer would be (-3.5, 15)

Answer 18

You're overcomplicating things.

(1/3)x + (1/5)(15) = 7
(1/3)x + 15/5 = 7
(1/3)x + 3 = 7 <15/5 =3>
(1/3)x = 4
x = 12 <multiply both sides by 3>

So your answer is (3,15)

Answer 19

I gotta go. Good job! For the record using substitution is usually much faster and easier than elimination. Solve for x in the equation and plug it into the second equation. As follows:

Equation 1:
1/3x+1/5y =7
(1/3)x = 7- (1/5)y
x = 21 - (3/5)y

1/6x–2/5y =-4
(1/6)(21 - (3/5)y) - (2/5)y = -4

Which looks pretty complicated but it's much more straightforward than the other method.

Answer 20

Sorry! the answer is (12,15) not (3,15). I'm tired.

Answer 21

Thank you very much :D