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2 years ago

How many solutions does the system have? y=3x-1 y=-2x+4

Mathematics

2 answers:

guapka [62]2 years ago

8 0

Answer:

This system has one answer.

Step-by-step explanation:

This system only has one answer which are: x=1 y=2

Hope this helped! :)

Vladimir79 [104]2 years ago

8 0

Answer:

(y,x) =(14,5)

Step-by-step explanation:

im in 10th grade

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I will give 10 pt to who ever answer the question

kap26 [50]

OK, because it is asking you to subtract a (negative) number and two negatives make a positive, the question would be -56 + 27.07 = ? with some simple addition, the answer would be <span>-28.93.

I hope this helps :)</span>

3 0

2 years ago

Read 2 more answers

2x + 8y = 5<br> 24x – 4y = –15

slega [8]

2x + 8y = 5

2/ 24x - 4y = -15

2x + 8y = 5

48x - 8x = -30

--------------------------

50x = -25

x = -1/2

2 * -1/2 + 8y = 5

-1 + 8y = 5

8y = 6

y = 6/8

y = 3/4

x = -1/2 y = 3/4

5 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

$\therefore \: x = 5$