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3 years ago

How do I use elimination process when solving two variable equations?

1 answer:

7 0

Solve simultaneously:

<span>15x-8y=52 and
6x+14y=38
</span>
Let's solve this system thru elimination by addition/subtraction:

Let's get rid of the x terms.
To do this, mult. the first eqn. by -6 and the second by 15. (This is not the only way in which we can do this!!) We get

-90x + 48y = -312
90x +210y = 570
--------------------------
Add all 3 columns together. We end up with:

258y = 258, so y = 1. You can find x by subst. 1 for y in either of the given equations.

If you'd like more help, please be specific about what you want.

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kvv77 [185]

Answer:

City 1 and City 2

Step-by-step explanation:

The water is moving down stream, moving from City 1-2-3-4 in that order.

if water was taken a the X mark, the pollutants would have to have come from the cities above where the sample was taken.

Hope this helps!

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3 years ago

Read 2 more answers

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ivann1987 [24]

Answer:

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Step-by-step explanation:

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4 years ago

Read 2 more answers

Graduation is 3 years away and you want to have $850 available for a trip. If your bank is offering a 3-year CD (certificate of

irina1246 [14]

Answer:

We need to put $801.88 amount in the bank.

Step-by-step explanation:

given that

t=3 yr

Need amount $850 after 3 yr so P= $850

Interest rate=2%

We know that for simple interest

$P=A\left(1+\dfrac{rt}{100}\right)$

Where r is the Interest rate,t is the time,A is the present amount and P is principle amount after t time.

Here given that P= $850

So now putting the values

$850=A\left(1+\dfrac{2\times 3}{100}\right)$

So A=$801.88

We need to put $801.88 amount in the bank.

7 0

4 years ago

I NEED HELP PLS THIS IS DUE IN 3 HOURS

Mariulka [41]

Answer:

Part 1) $x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})$

Part 2) $x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})$

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

Part 1)

in this problem we have

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

$a=1\\b=-2\\c=-2$

substitute in the formula

$x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-2)}} {2(1)}\\\\x=\frac{2(+/-)\sqrt{12}} {2}\\\\x=\frac{2(+/-)2\sqrt{3}} {2}\\\\x_1=\frac{2(+)2\sqrt{3}} {2}=1+\sqrt{3}\\\\x_2=\frac{2(-)2\sqrt{3}} {2}=1-\sqrt{3}$