# Thread: Solving by elimination or subsitution.

1. ## Solving by elimination or subsitution.

Hey guys it has been so long since I have done a lot of algebra and I am working on this take home assignment and am trying hard to figure out how to tackle some of these.

The instructions state: "Solve the following systems of equations by elimination, substitution, or graphically.

A 3x-y=6 & x+4y=3 B y+2=-x^2+1 & x2+1=2y+6 C x-2y+3z=10 & 2x-3y-z=8 & 5x-9y+8z=20

For A I got X being 2 1/3 and Y being 3. I am struggling on the other two. Any help is appreciated.

2. B)
$\displaystyle \left\{\begin{array}{ll}y+2=-x^2+1\\2y+6=x^2+1\end{array}\right.$
By elimination:
Adding the equations we'll eliminate $\displaystyle x^2$ and we get
$\displaystyle 3y+8=2\Rightarrow 3y=-6\Rightarrow y=-2$
Now, plug y in the first (or second) equation and find x.

By substitution:
From the first equation $\displaystyle y=-x^2-1$.
Substitute y in the second equation and solve for x.

Graphically:

3. C)
$\displaystyle \left\{\begin{array}{lll}x-2y+3z=10\\2x-3y-z=8\\5z-9y+8z=20\end{array}\right.$
By elimination:
We'll eliminate z:
Multiply the second equation by 3 and add to the first:
$\displaystyle 7x-11y=21$ (1)
Multiply the second by 8 and add to rhe third:
$\displaystyle 21x-33y=84$ (2)
Now, multiply (1) by -3 and add to (2):
$\displaystyle 0=21$, which is obviously false. So the system has no solution.

By substitution:
From the first equation $\displaystyle x=2y-3z+10$
Substitute x in the others two equations and you'll obtain a system in y and z. Apply the same method to this system.

4. Originally Posted by Ty Durdan
...

The instructions state: "Solve the following systems of equations by elimination, substitution, or graphically.

A 3x-y=6 & x+4y=3 ...

For A I got X being 2 1/3 and Y being 3.
Hello,

I don't know how you did A) - but:

$\displaystyle \left\{\begin{array}{ll}3x-y=6 \\ x+4y=3 \end{array}\right.$ . Multiply the 1rst equation by 4. You'll get:

$\displaystyle \left\{\begin{array}{ll}12x-4y=24 \\ x+4y=3 \end{array}\right.$ and then add columnwise(?):

$\displaystyle 13x = 27~\Longleftrightarrow~ \boxed{x=\frac{27}{13}}$. Plug in this value into one of the equations to calculate y:

$\displaystyle 3x-y=6~\Longrightarrow~3 \cdot \frac{27}{13}-y=6~\Longrightarrow~ y=\frac{81}{13}-\frac{78}{13}=\boxed{\frac{3}{13}}$