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Algebra 1 : Systems of Equations

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Example Questions

Example Question #61 : Systems Of Equations

Solve for : 3x(x+4)=x+20

Possible Answers:

$\left \{ -15, 4\right \}$

$\left \{ -4, 15\right \}$

$\left \{ -20,-4\right \}$

$\left \{ -\frac{4}{3},5\right \}$

$\left \{ -5, \frac{4}{3}\right \}$

Correct answer:

$\left \{ -5, \frac{4}{3}\right \}$

Explanation:

Eliminate parentheses, then write this quadratic equation in standard form, with all nonzero terms on one side:

3x(x+4)=x+20

$3x \cdot x +3x \cdot 4=x+20$

$3x^{2} +12x=x+20$

$3x^{2} +12x-x-20 =0$

$3x^{2} +11x-20 =0$

Now factor the quadratic expression using the -method - split the middle term into two terms whose coefficients add up to 11 and have product $3 \cdot (-20) = -60$ . These numbers are 15,-4

$3x^{2} -4x+15x-20 =0$

x (3x -4) + 5(3x -4)=0

(x + 5)(3x -4) =0

Set each factor to 0:

x+5 = 0

x = -5

3x-4 = 0

3x=4

$x = \frac{4}{3}$

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Example Question #1 : Quadratic Equations

$Solve\;for\;x,\;(x-3)^2 =9$

Possible Answers:

6,12

$\fn_cm 0,6$

Correct answer:

$\fn_cm 0,6$

Explanation:

$x-3 = 3 \;or \;x-3 = -3$

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Example Question #10 : How To Find The Solution To A Quadratic Equation

$Solve \;for\;p,(p-5)(p+3)=0$

Possible Answers:

-5,3

5,-3

-5,-3

5,3

Correct answer:

5,-3

Explanation:

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Example Question #11 : Quadratic Equations

$Solve \;for\;q,q^2+9q+14=0$

Possible Answers:

14,1

7,2

7,-2

-7,2

-7,-2

Correct answer:

-7,-2

Explanation:

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Example Question #61 : Systems Of Equations

$Solve \; for\;t,t^2 + 6t + 8 = 0$

Possible Answers:

t = -1, -8

t = -4, -2

t= 6, 8

t= 4, 2

Correct answer:

t = -4, -2

Explanation:

$Factor \;t^2 + 6t + 8 = 0$

(t+4)(t+2)=0

t=-4,-2

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Example Question #62 : Systems Of Equations

$Solve\;for\;x,x^2 + 7x + 15 = 5$

Possible Answers:

x= -3,-5

$\fn_cm x= -2, -5$

x= 3,5

x= 2, 5

Correct answer:

$\fn_cm x= -2, -5$

Explanation:

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Example Question #1 : Quadratic Equations

Solve for $\small x$ .

$\small x^2+6x+8=0$

Possible Answers:

$\small x=-2\ or\ x=-4$

$\small x=-2\ or\ x=4$

$\small x=2\ or\ x=4$

$\small x=2\ or\ x=-4$

Correct answer:

$\small x=-2\ or\ x=-4$

Explanation:

$\small x^2+6x+8=0$

Solve by factoring. We need to find two factors that multiply to eight and add to six.

$\small 2*4=8\ \text{and}\ 2+4=6$

$\small x^2+6x+8=(x+2)(x+4)=0$

One of these factors must equal zero in order for the equation to be true.

$\small x+2=0\ or\ x+4=0$

$\small x=-2\ or\ x=-4$

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Example Question #1 : Quadratic Equations

Solve $3x^{2}-12=0$

Possible Answers:

x=3, 2

x=-3, 2

x=2, -2

x=4, -4

x=-3, 12

Correct answer:

x=2, -2

Explanation:

First, we must factor out any common factors between the two terms. Both 3 and 12 share a factor of 3, so we can "take" 3 out, like this:

$3(x^{2}-4)$ .

Inside the parentheses, it becomes clear that this is a difference of squares problem (a special factor), which can be solved with the equation

$a^{2}-b^{2}=(a-b)(a+b)$ .

Thus, $x^{2}-2^{2}=(x-2)(x+2)$ .

Now, we can set each factor to 0, and solve for :

x-2=0

x=2

and

x+2=0

x=-2 .

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Example Question #15 : Quadratic Equations

The product between an integer and another integer that's units greater is equal to . Find these integers.

Possible Answers:

none of these

$-1\ and\ 2$

$-2\ and\ 2$

$-3\ and\ 1$

$2\ and\ 4$

Correct answer:

$-2\ and\ 2$

Explanation:

We'll call one of the integers , which makes the other integer x+4 . The product between and x+4 is . This is written as

x(x+4)=-4

Solve for .

x^2+4x=-4

x^2+4x+4=0

We can solve this either by using the quadratic formula or by factoring. Factoring seems to be the quickest method.

(x+2)(x+2)=0

(x+2)^2=0

Take the square root of both sides.

$\sqrt{(x+2)^2}=\sqrt{0}$

x+2=0

x=-2

The first integer is , which we can use to find the other integer by using x+4 .

x+4=-2+4=2

The second integer is 2.

Therefore, the two integers are and .

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Example Question #2322 : Algebra 1

Find the solutions:

x^2 - 3x - 10 = 0

Possible Answers:

x = 1, 10

x = -5, -2

x = -5, 2

x = 5, 2

x = 5, -2

Correct answer:

x = 5, -2

Explanation:

Since the trinomial starts with x^2 , the first term in each of the binomials will be :

x^2 - 3x - 10 = 0

$(x\quad)(x\quad) = 0$

Since you have a at the end of the trinomial, the two terms at the end of the two binomials must multiply to . Since it is a negative , the signs in the two binomials will be opposite.

and work:

(x - 5)(x + 2) = 0

Now you set each binomial equal to independently, and solve for .

$x - 5 = 0 \qquad x + 2 = 0$

$x = 5 \qquad x= -2$

x = 5, -2

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Algebra 1 : Systems of Equations

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #61 : Systems Of Equations

Example Question #1 : Quadratic Equations

Example Question #10 : How To Find The Solution To A Quadratic Equation

Example Question #11 : Quadratic Equations

Example Question #61 : Systems Of Equations

Example Question #62 : Systems Of Equations

Example Question #1 : Quadratic Equations

Example Question #1 : Quadratic Equations

Example Question #15 : Quadratic Equations

Example Question #2322 : Algebra 1

All Algebra 1 Resources

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