Algebra II System of Equations Question

Rather than doing them for you, I'll guide you through problem number three and have you do the rest. Problem one may look tough, but it's rather simple if you know properties of exponents and logs.

Alrighty, problem number three:

\(x + y = -2\)

\(y+4x=x^2\)

So, using substitution, we want to get one of the equations to start as either y= or x=. I'll just pick the first equation and make it start with  y=.

\(x +y = -2 \Rightarrow x + y -x = -2 - x \Rightarrow y = -2-x\)

So, now our system looks like this:

\(y=-2-x\)

\(y+4x=x^2\)

Now let's work on the second equation in our system. Since we know that y = -2 - x, we can just plug that in whenever we see a y in equation two:

\(y+4x = x^2 \Rightarrow (-2-x) + 4x = x^2 \)

Woohoo! All x's! Let's simplify this to get our x-values now.

\(-2-x+4x = x^2 \Rightarrow -2 + 3x = x^2 \Rightarrow 0 = x^2-3x+2\) -->Now we factor

\(0 = x^2 -3x +2 \Rightarrow 0 = (x-2)(x-1) \Rightarrow 0 = x-2, 0 = x-1 \Rightarrow x=2, x=1\)

So, we've got our two x-values! Neat! Now let's go back to equation 1 and find our y-values using those nifty x-values we just found.

\(y = -2 -x\Rightarrow y = -2 - 2, y = -2 - 1 \Rightarrow y= -4, y = -3\)

So, we've got two x-values and two y-values; now we just put them together and we've got two nice ordered pairs as answers!

Pair 1: (2, -4)

Pair 2: (1, -3)

The other 2 follow the same process as this, just some different algebra to find your values.

1.

y-9=e2x     Add  9 to both sides     y  =  e^(2x)  + 9

3=y-7ex      Subtract  7e^x   to both sides   y  =  3 + 7e^x

Set the y's equal

e^(2x)  + 9   =   3 +  7e^x    rearrange as

e^(2x)  - 7e^x  +  6  =  0    factor

(e^x  - 6)(e^x - 1)   =  0        set each factor to 0   and solve

e^x  -  6  = 0

e^x    = 6          take the Ln of both sides

Ln e^x  = Ln 6    and we can write

x * Ln e   =  Ln 6

x =   Ln 6

Using a similar proceedure

x  = Ln 1    = 0

So.....the two solutions are   x  = Ln 6    and   x  = 0