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3 years ago

Solve the system of equations algebraically. Show all of your steps. y=x^2+2x y=3x+20

Mathematics

1 answer:

alexira [117]3 years ago

6 0

Answer:

I'm pretty sure its y = x^2 + 5x + 20

Step-by-step explanation:

This is because i added them (or combined)

Like this

x^2 + 2x

+ 3x + 20

________

x^2 + 5x + 20

Please Correct Me If I Am Wrong Okay love.

Again sorry if i'm wrong.

I'm pretty sure i'm not.

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What is the value of the discriminant, b2 − 4ac, for the quadratic equation 0 = x2 − 4x + 5, and what does it mean about the num

forsale [732]

(-4)^2-4(1)(5)
16-4(1)(5)
16-4(5)
16-20
-4

The discriminant is negative so there are no real number solutions.

3 0

3 years ago

Read 2 more answers

In converting 30 pounds to ounces, what unit (omit the number) would you

gladu [14]

Step-by-step explanation:

to convert pounds into ounces you'd

actually multiply 16 by 30:

30 * 16

you're doing this because if one pound equals 16 ounces, if you have 30 pounds then you must've had more ounces. therefore you'd multiply.

which is 480 pounds

Enjoy!

6 0

3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

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

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

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

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

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$