The solution of the system can be x - 3y = 4 only if both the equations can be simplified to x - 3y = 4.
This will mean that both the equations will result in the same line which is x - 3y = 4 and thus have infinitely many solutions.
Second equation is:
Qx - 6y = 8
Taking 2 common we get:
(Q/2)x - 3y = 4
Comparing this equation to x- 3y = 4, we can say that
Q/2 = 1
So,
Q = 2
Therefore, the second equation will be:
2x - 6y = 8
<span>3x^2y^2 − 2xy^2 − 8y^2 =
y^2 (3x^2 - 2x - 8) =
factoring with leading coefficient:
for ax2+bx+c find two numbers n,m, that m*n = a*c and m+n = b
</span><span><span>
3x^2 - 2x - 8
a=3, b=-2, c=-8
</span>a*c = 3*(-8) = -24
-24=(-6)*4 and -6+4=-2, so m=-6 and n=4
replace bx with mx + nx and factor by grouping
</span><span>
3x^2 - 2x - 8 = </span>3x^2 -6x + 4x -8 = 3x(x-2) + 4(x-2) = (3x+4)(x-2)
answer:
<span>3x^2y^2 − 2xy^2 − 8y^2 = y^2(3x+4)(x-2)</span>
y = x - 3
1. Subtract 7 from the right to the left.
* 4y = 4x - 12
2. Divide the 4 (constant) next to the 'y' to the other side.
* y = 4/4x - 12/4
* y = x - 3
I believe this is a function. I am not sure because one of the pairs only has a y value.
But the rule is there can only be one x value for one y value so for example, you can have (3.7) and (3,9) and the same for the y values.
