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const2013 [10]
2 years ago
10

Solve the system of equations with substitution or elimination method. please help 2x+y=4 and 6x+7y=12

Mathematics
2 answers:
GuDViN [60]2 years ago
8 0

Answer:

x=2-y/2

y=12/7-6x/7

Step-by-step explanation:

klemol [59]2 years ago
4 0
I solved using elimination, multiplying the first equation by -3. My work is shown in the attached image.
x=2
y=0

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3 years ago
What is <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B7%7D%20%20%3D%20%20%5Cfrac%7B4%7D%7B5%7D%20" id="TexFormula1"
frosja888 [35]

Multply both sides of the equation by 7.

\dfrac{x}{7}\cdot 7=\dfrac{4}{5}\cdot 7\\\\x=\dfrac{28}{5}=5.6

_____

When the variable is in the numerator of a proportion like this, you can solve it by multiplying by that denominator. That multiplication cancels the denominator and gives you the value of the variable.

In general, you multiply by the reciprocal of the coefficient of x. When that coefficient is 1/7, you multiply by 7/1. Of course, you know that 7/7 = 1 and that x/1 = x.

7 0
3 years ago
X and y are positive integers. explain why incorrect to claim that
dedylja [7]

You can find counterexamples to disprove this claim. We have positive integers that are perfect square numbers; when we take the square root of those numbers, we get an integer.

For example, the square root of 1 is 1, which is an integer. So if y = 1, then the denominator becomes an integer and thus we get a quotient of two integers (since x is also defined to be an integer), the definition of a rational number.

Example: x = 2, y = 1 ends up with \frac{2}{\sqrt{1}} = \frac{2}{1} which is rational. This goes against the claim that x/\sqrt{y} is always irrational for positive integers x and y.

Any integer y that is a perfect square will work to disprove this claim, e.g. y = 1, y = 4, y= 9, y = 16. So it is not always irrational.

4 0
3 years ago
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Marrrta [24]
C
step by step explanation:
8 0
3 years ago
6. The population of a town is 680 000 correct to the nearest 10 000. Write down
serg [7]

Answer:

a) 675 000

b) 685 000

Step-by-step explanation:

The population of a town is 680 000 correct to the nearest 10 000.

a) To find it lower bound, we level of accuracy by 2 and then subtract from 680 000

The lower bound is:

680 000-5000=675,000

Therefore the least possible population of the town is 675 000

b) We repeat the same process to find the upper bound

680 000+5000=685,000

6 0
3 years ago
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