# Math Problems for 9 – 12 – 16

## Problem 1

When factored completely, x3 – 13x2 – 30x is
(1) x(x + 3)(x – 10)
(2) x(x – 3)(x – 10)
(3) x(x + 2)(x – 15)
(4) x(x – 2 )(x + 15)

Cuando se factoriza por completo, x3 – 13x2 – 30x equivale a
(1) x(x + 3)(x – 10)
(2) x(x – 3)(x – 10)
(3) x(x + 2)(x – 15)
(4) x(x – 2 )(x + 15)

Discussion and Solution:
Since each element in the original equation contains an x, we can factor an x out of each, yielding:

x(x2 – 13x – 30)

We next need to factor the parenthetical expression, arriving at the final solution.

x(x2 – 13x – 30) = x(x + 2)(x – 15)

Choice (3).

## Problem 2

When (2x – 3)2 is subtracted from 5x2, the result is
(1) x2 – 12x – 9
(2) x2 – 12x + 9
(3) x2 + 12x – 9
(4) x2 + 12x + 9

Cuando (2x – 3)2 se resta de 5x2, el resultado es
(1) x2 – 12x – 9
(2) x2 – 12x + 9
(3) x2 + 12x – 9
(4) x2 + 12x + 9

Discussion and Solution:
Watch the English. It is asking for 5x2 – (2x – 3)2, not (2x – 3)2 – 5x2.
We first expand (2x – 3)2, getting

5x2 – (4x2 – 12x + 9)

We then negate this because of the negative sign in front of it, getting

5x2 – 4x2 + 12x – 9

We then combine like terms, getting

x2 + 12x – 9

Choice (3).

## Problem 3

Which equation is equivalent to y – 34 = x(x – 12)?
(1) y = (x – 17)(x + 2)
(2) y = (x – 17)(x – 2)
(3) y = (x – 6)2 + 2
(4) y = (x – 6)2 – 2

¿Qué ecuación es equivalente a y – 34 = x(x – 12)?
(1) y = (x – 17)(x + 2)
(2) y = (x – 17)(x – 2)
(3) y = (x – 6)2 + 2
(4) y = (x – 6)2 – 2

Discussion and Solution:
First we remove the parentheses, getting

y – 34 = x2 – 12x

We next add 36 to both sides, getting

y – 34 + 36 = x2 – 12x + 36

This factors into

y + 2 = (x – 6)2

Finally, we subtract 2 from each side, getting

y = (x – 6)2 – 2

Choice (4).

## Problem 4

Which pair of equations could not be used to solve the following equations for x and y?

4x + 2y = 22

– 2x + 2y = – 8

(1) 4x + 2y = 22
2x – 2y = 8

(2) 4x + 2y = 22
– 4x + 4y = – 16

(3) 12x + 6y = 66
6x – 6y = 24

(4) 8x + 4y = 44
– 8x + 8y = – 8

¿Qué par de ecuaciones no podría usarse para resolver las siguientes ecuaciones para x e y?

4x + 2y = 22

– 2x + 2y = – 8

(1) 4x + 2y = 22
2x – 2y = 8

(2) 4x + 2y = 22
– 4x + 4y = – 16

(3) 12x + 6y = 66
6x – 6y = 24

(4) 8x + 4y = 44
– 8x + 8y = – 8

Discussion and Solution:
Remember, there are three ways to solve two equations with two unknowns –

• Graph each equation on the same set of axes. Where they intersect, if they do, is the solution set;
• Solve one of the equations for one of the unknowns. Then substitute this equality into the other equation and solve. Once the second unknown is determined, substitute this value into either of the original equations to determine the second unknown; and
• Multiply one (or both) of the two original equations by a small whole number and then add or subtract the two equations, thereby eliminating one of the original unknowns. Substitute this value into either of the two original equations, solving for the second unknown.

From the looks of our choices, we multiply the first equation by 2 and the second equation by 4. When we add these two resulting equations, we would eliminate the value of x.

Choice (4).

## Problem 5

Solve the equation for y.

(y – 3)2 = 4y – 12

Resuelva la ecuación para y.

(y – 3)2 = 4y – 12

Discussion and Solution:

First we eliminate the parentheses:

(y – 3)2 = 4y – 12

y2 – 6y + 9 = 4y – 12

We next move all the terms to one side of the equation, getting

y2 – 6y + 9 – 4y + 12 = 0

Combine like terms.

y2 – 10y + 21 = 0

Factor this equation.

(y – 7)(y – 3) = 0

Set each factor equal to 0 and solve for y.

y – 7 = 0

y = 7

and

y – 3 = 0

y = 3

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