# Thread: A few Algebra 2 problems

1. ## A few Algebra 2 problems

Problem one:
I have tried this over and over again but i keep getting answers like 12=12 in the end. Solve the system of equations:
6x-10y=12
-15x+25y=-30

Problem two:
I never really learned this before. Factor:
16x²y²+22xy-3

Problem three:
Never really learned this either. Graph:
y=2x²+5x-12

Thanks for the help in advance =)

2. Originally Posted by nike23
Problem one:
I have tried this over and over again but i keep getting answers like 12=12 in the end. Solve the system of equations:
6x-10y=12
-15x+25y=-30

Problem two:
I never really learned this before. Factor:
16x²y²+22xy-3

Problem three:
Never really learned this either. Graph:
y=2x²+5x-12

Thanks for the help in advance =)
1. Immediately I see that I can cancel equation 2 by a factor of 5:

$-15x+25y = 30 \rightarrow -3x+5y = 6$

If I multiply this equation by 2 I get equation 1 which would lead to the problem you're getting. It means there are infinite solutions

2. note that $(ab)^2 = a^2b^2$ and $16x^2y^2 = (4xy)^2$

Let $xy = u$ and it reduces to $16u^2 + 22u -3$ which should be easier to factorise

3. Find out where x=0 and where y=0. For where x=0 solve f(0) and this will usually be the lowest point. To find where y=0 set f(x) to 0 and solve. This will give the x-intercepts.
Using these three points sketch the graph

3. 6x-10y=12
-15x+25y=-30

$x+ \frac {-25}{15y} = 2$

$x= 2+ \frac {25}{15y}$

$6(2+ \frac {25}{15y}) -10y = 12$

$12+ \frac {150}{15y} -10y =12$

$\frac {150}{15y} - 10y = 0$

$150-150y^2 = 0$

$150y^2 = -150$

go on!

4. Originally Posted by jgv115
6x-10y=12
-15x+25y=-30

$x+ \frac {-25}{15y} = 2$

$x= 2+ \frac {25}{15y}$

$6(2+ \frac {25}{15y}) -10y = 12$

$12+ \frac {150}{15y} -10y =12$

$\frac {150}{15y} - 10y = 0$

$150-150y^2 = 0$

$150y^2 = -150$

go on!
If you "go on," as you say, you will get $y^2 = -1$, which makes no sense. I suspect that things went wrong when you went from
this step:
$\frac {150}{15y} - 10y = 0$
to this step:
$150-150y^2 = 0$

You multiplied both sides by 15y. Doing so introduces extraneous solutions, so you shouldn't have done that. e^(i*pi)'s method is the better one for this problem.

Originally Posted by e^(i*pi)
3. Find out where x=0 and where y=0. For where x=0 solve f(0) and this will usually be the lowest point. To find where y=0 set f(x) to 0 and solve. This will give the x-intercepts.
Using these three points sketch the graph
I'm not sure what you mean by "lowest point." That sounds like vertex to me (assuming the parabola opens upward), but the vertex is not at (0, -12), as you can see from the graph.

nike23: While it's a good idea to find the x- and y-intercepts, I'd also find the vertex. You have to put the equation into vertex form:
$y = a(x - h)^2 + k$
where (h, k) is the vertex. To do this, you'll have to complete the square. If you don't know how to do that, you can remember these formulas instead: if you have a quadratic equation $f(x) = ax^2 + bx + c$, the vertex (h, k) is found by the following...
$h = -\frac{b}{2a}$
$k = c - ah^2 \;\text{or}\; f(h)$.

So, for $f(x) = 2x^2 + 5x - 12$, a = 2, b = 5, and c = -12, so
$h = -\frac{5}{2(2)} = -\frac{5}{4}$ and
$k = c - ah^2 = -12 - 2\left(-\frac{5}{4}\right)^2 = -12 - 2\left(\frac{25}{16}\right) = -12 - \frac{25}{8} = -\frac{121}{8}$.

So the vertex is at $\left(-\frac{5}{4},\;-\frac{121}{8}\right)$.

01

5. Originally Posted by jgv115
6x-10y=12
-15x+25y=-30

$x+ \frac {-25}{15y} = 2$

$x= 2+ \frac {25}{15y}$

$6(2+ \frac {25}{15y}) -10y = 12$

$12+ \frac {150}{15y} -10y =12$

$\frac {150}{15y} - 10y = 0$

$150-150y^2 = 0$

$150y^2 = -150$

go on!
It should be $150y^2 = 150$