y=(x+a)(x+b) intersects the x-axis
so y=0
(x+a)(x+b)=0
x=-a,-b
1) a+b=-1, from this we can't find a,b
INSUFF

2)The graph intersects the Y-axis at (0, -6)
y=(x+a)(x+b)
put the vlues
-6=ab, again we can't find a and b
INSUFF
combine
a+b=-1
ab=-6
a=2,b=-3 or, a=-3,b=2
INSUFF

combining both
point X=-2,3
SUFF

so C

Answer is correct, but the points should be (-2, 0) & (3,0)

Combining 1) and 2)

Y = x^2 +xb+ax+ab
= x^2 + x(b+a)+ab
= x^2 – x -6
=(x-3) (x+2)
X = 3 or x = -2

navalpike wrote:Answer is correct, but the points should be (-2, 0) & (3,0)

Combining 1) and 2)

Y = x^2 +xb+ax+ab
= x^2 + x(b+a)+ab
= x^2 – x -6
=(x-3) (x+2)
X = 3 or x = -2

oh yeah, I just gave you the X coordinates. Y-coordinates will obviously be zero because the function intersects x axis at those points

anyway to solve this via the system of equations looking at number of equations and number of unknowns? thx

abcdefg wrote:In the xy-plane, at what point does the graph of y=(x+a)(x+b) intersect the x-axis?

1. a+b=-1
2. The graph intersects the Y-axis at (0, -6)


The OA is C . I'm wondering if anyone can help me get an answer via the systems of equations method because it'd be alot fasther than solving the equations manually.

The original equation becomes y = x^2 +ax +bx +ab
If we simplify statements 1 and 2, we get 2 more equations:

a + b = -1
ab = -6

Combining them we have 3 equations but 4 unknowns (y,x, a,b). How can the answer C be justified via systems of equations? Thanks.

To go back to your actual question, you forgot to add the equation y=0 to your list (we know that y=0 since we're looking for the point at which the line intersects the x-axis). However, we have a quadratic, so we don't satisfy the rule, which states:

To solve for a system of n variables, one requires n distinct linear equations.

Since y=0, we basically have the quadratic:

x^2 + (a+b)x +ab = 0

Substituting in, we get:

x^2 -x -6 = 0
(x-3)(x+2) = 0
x=3 or -2

HOWEVER, perhaps someone can alleviate my confusion, why is the answer C and not E? Don't we have two points of intersection and therefore more than 1 answer to the question?

Did the original question ask "at what point" or "at what points"? If it's the former, the answer should be E; if it's the latter, the answer should be C.

I believe that the question is from GMAT prep and has been misstated. I found another link for this question with different wording.

https://www.beatthegmat.com/problem-from ... 13502.html

But you are correct (as usual) . As worded, the answer should have been E. My mistake.

When you say a linear equation, can the equation be with inequalities?

For example"
8x > 4y + 6x
6 – 5x > -13y

would these 2 work to give us unique answers?

abcdefg wrote:When you say a linear equation, can the equation be with inequalities?

For example"
8x > 4y + 6x
6 – 5x > -13y

would these 2 work to give us unique answers?

No, they need to be linear equations; inequalites are... well... inequalities!