If you have at least as many equations as your have unknown variables, the 'system' is solvable (unless the equations are copies of eachother).
In this case, isolate one letter and plug it in the other. I'm going for the 2a in the bottom one (it doesn't matter)
2a - 4b = 12 => (divide by two and move the b's to the other side)
a = 6 + 2b. (plug this one into the top equation)
4(6+2b) + 6b = 10 => 24+8b+6b = 10 => 24 + 14b = 10 => 14b = -14 => b=-1
a = 6 + -2 = 4.
So a=4 and b=-1.
f(x) being even means
f(x) = f(-x)
So the zeros come in positive and negative pairs. If there are an odd number of intercepts like there are here, it's because one of them is x=0 which is its own negation.
Given zero x=6 we know x=-6 is also a zero.
So we know three zeros, and know the other two zeros are a positive and negative pair.
The only choice with (-6,0) and (0,0) is A.
Choice A
To evaluate the expression all you have to do is substitute the value for the variable.
3 + y + 6, y = 5
3 + 5 + 6
8 + 6
14.
The solution is 14.
It is answer choice A because you have to take the the GCF of each number which is 2x. Then if you were to divide the numbers by 2x you would put those numbers in the brackets.
because
and
and
Move all terms to one sides
3x^2 - 14x - 5 = 0
Split the second term in 3x^2 - 14x - 5 into two terms
3x^2 + x - 15x - 5 = 0
Factor out the common terms in the first two terms, then in the last two terms;
x(3x + 1) - 5(3x + 1) = 0
Factor out the common term 3x + 1
(3x + 1)(x - 5) = 0
Solve for x;
<u>x = -1/3, 5</u>
