# Thread: Can you help me out with some substitution

1. ## Can you help me out with some substitution

Going through my substitution section of my homework, and I successfully did most of them but had trouble with two of them:

Determine the intersection of the following systems:
a) y = 2x – 7 and y = - x2 -4 x – 3

Heres my attempt:

(2x-7)=-x2 -4x-3

0=x2 -4x-3-2x+7

-1(0=-x2 -6x+4)

0=x2+6x-4

In every other question we've had we could just simplify this last step to end up with an answer like (x-x)(x-x), however with this one I can't see what multiplies to -4, and adds to 6

Then there's the second one
b) x2 + y2 = 30 and 2 x – 3 y = 10

2x-3y=10 ~ y= 2x + 10
3 3
x2 +(2x + 10)2 = 30
.........3......3 [sorry about dots, still learning how to type in equations]

Is this correct so far? It doesn't feel like it is....

2. ## Re: Can you help me out with some substitution

Originally Posted by Urpes
Going through my substitution section of my homework, and I successfully did most of them but had trouble with two of them: Determine the intersection of the following systems:
a) $y = 2x - 7$ and $y = - x^2 -4 x - 3$
Heres my attempt:

$0=x^2+6x-4$
The equation $x^2+6x-4=0$ cannot be solved by factoring.
Do you know the quadratic formula?

3. ## Re: Can you help me out with some substitution

Originally Posted by Plato
The equation $x^2+6x-4=0$ cannot be solved by factoring.
Do you know the quadratic formula?
Yes I do, but I never used it after substitution but heres my attempt :

I'm not even sure how to start the second one though.

(better look at second one)

This is going to be on my test tomorrow and I could really use the help

4. ## Re: Can you help me out with some substitution

The quad. formula is correct. However in step 7 you used a=6 when a=1. Also, steps 8 and onward, you forgot that it's '-b' i.e. -6 not 6

5. ## Re: Can you help me out with some substitution

For quadratic equations you can find the zeros by the following methods (in order of decreasing difficulty): Rational Root Theorem, Factoring, Completing the Square, Quadratic Equation.

It's useful to know all 4.
For problem 1, completing the square you arrive at +/- sqrt(13)-3

Problem 2, you seem to be doing it right. Squaring that linear expression gives you a trinomial of the 2nd degree. Combine terms. Use quadratic equations if it's messy.

6. ## Re: Can you help me out with some substitution

Originally Posted by Urpes
Going through my substitution section of my homework, and I successfully did most of them but had trouble with two of them:

Determine the intersection of the following systems:
a) y = 2x – 7 and y = - x2 -4 x – 3

Heres my attempt:

(2x-7)=-x2 -4x-3

0=x2 -4x-3-2x+7

-1(0=-x2 -6x+4)

0=x2+6x-4

In every other question we've had we could just simplify this last step to end up with an answer like (x-x)(x-x), however with this one I can't see what multiplies to -4, and adds to 6

Then there's the second one
b) x2 + y2 = 30 and 2 x – 3 y = 10

2x-3y=10 ~ y= 2x + 10
3 3
x2 +(2x + 10)2 = 30
.........3......3 [sorry about dots, still learning how to type in equations]

Is this correct so far? It doesn't feel like it is....
Please refer to our LaTeX forum. Basic use is quick and much easier for us to understand what equations you are trying to type. Look up \sqrt and \frac.

-Dan

7. ## Re: Can you help me out with some substitution

Originally Posted by Urpes
Going through my substitution section of my homework, and I successfully did most of them but had trouble with two of them:

Determine the intersection of the following systems:
a) y = 2x – 7 and y = - x2 -4 x – 3

Heres my attempt:

(2x-7)=-x2 -4x-3

0=-x2 -4x-3-2x+7 Where did your negative disappear to?

-1(0=-x2 -6x+4)

0=x2+6x-4

In every other question we've had we could just simplify this last step to end up with an answer like (x-x)(x-x) Really, you managed to get 0 times 0 in each case?, however with this one I can't see what multiplies to -4, and adds to 6

Then there's the second one
b) x2 + y2 = 30 and 2 x – 3 y = 10

2x-3y=10 ~ y= 2x + 10
3 3
x2 +(2x + 10)2 = 30
.........3......3 [sorry about dots, still learning how to type in equations]

Is this correct so far? It doesn't feel like it is....
Your original transposition in the second equation is incorrect.

\displaystyle \begin{align*} 2x - 3y &= 10 \\ 2x &= 3y + 10 \\ 2x - 10 &= 3y \\ \frac{2}{3} x - \frac{10}{3} &= y \end{align*}

Your approach would otherwise be correct, so now expand out the brackets and collect your like terms, then solve the resulting quadratic equation for x.

8. ## Re: Can you help me out with some substitution

topsquark I will check that out tomorrow, I have to go to bed right now, so I'm not going to be able to do it now.

For now I'll just clean it up by posting a picture, hopefully if someone answers I can check it in the morning before my test.

Can anyone solve this for me? This is a sample question for my quiz tomorrow that I'm still not ready for, could really, really use an answer.

9. ## Re: Can you help me out with some substitution

Originally Posted by Urpes
topsquark I will check that out tomorrow, I have to go to bed right now, so I'm not going to be able to do it now.

For now I'll just clean it up by posting a picture, hopefully if someone answers I can check it in the morning before my test.

Can anyone solve this for me? This is a sample question for my quiz tomorrow that I'm still not ready for, could really, really use an answer.
Did you even bother to read the post that I did JUST before this one?

10. ## Re: Can you help me out with some substitution

So I assume the next step would be:

If I sub in what you gave me into x2 + y2 = 30 I get the following

I don't know where to start to solve for x....

11. ## Re: Can you help me out with some substitution

I already told you what to do, expand your brackets, collect your like terms, set the equation equal to 0 and solve the quadratic.