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guajiro [1.7K]
3 years ago
7

Solve the following system of equations. 9x + 4y = 4 -5x + 7y = 7

Mathematics
2 answers:
enyata [817]3 years ago
7 0

Answer:

this is the answer with steps

hope it helps!

ale4655 [162]3 years ago
5 0

Answer:

The solution is:

(0, 1)

Step-by-step explanation:

We have the following equations

9x + 4y = 4

-5x + 7y = 7

To solve the system multiply by \frac{9}{5} the second equation and add it to the first equation

-5*\frac{9}{5}x + 7\frac{9}{5}y = 7\frac{9}{5}

-9x + \frac{63}{5}y = \frac{63}{5}

9x + 4y = 4

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

\frac{83}{5}y=\frac{83}{5}

y=1

Now substitute the value of y in any of the two equations and solve for x

9x + 4(1) = 4

9x +4 = 4

9x = 4-4

9x = 0

x=0

The solution is:

(0, 1)

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4 0
3 years ago
The weight of Jane and Jessica was in the ratio of 8 : 9. Jane gained 2 kg while Jessica lost 4 kg. They then had the same weigh
postnew [5]

Answer:

  48 kg

Step-by-step explanation:

The given relations can be used to write a system of equations for the two weights. Those can be solved to find Jane's weight.

__

<h3>setup</h3>

Let x and y represent Jane's and Jessica's original weight, respectively. The ratio of weights was ...

  x/y = 8/9

After the changes in weight, they were equal:

  x+2 = y-4

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<h3>solution</h3>

Adding 4 to the second equation, we have an expression for y that can be substituted into the first equation.

  y = x +6 . . . . . . . . . . solve the second equation for y

  x/(x+6) = 8/9 . . . . . . substitute for y in the first equation

  9x = 8(x +6) . . . . . . . multiply by 9(x+6)

  x = 48 . . . . . . . . . . . simplify and subtract 8x

Jane weighed 48 kg at first.

_____

<em>Alternate solution</em>

The original difference in "ratio units" was 9-8 = 1 ratio unit. We find that this corresponds to 6 kg after the weight changes make the weights equal. Then 8 ratio units will be 8(6 kg) = 48 kg—Jane's original weight.

(This mental solution is virtually the same as the solution using equations shown above.)

3 0
1 year ago
Which do you think is easier to understand and why do you think so, multiplying radicals or multiplying polynomials?
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I believe that its easier to understand the multiplication of a radical than that of a polynomial.

<h3>How to illustrate the information?</h3>

In mathematics, a radical is the opposite of an exponent that is represented with a symbol '√' also known as root.

A polynomial is an expression which is composed of variables, constants and exponents, that are combined using the mathematical operations such as addition, subtraction, multiplication and division.

When multiplying radicals with the same index, multiply under the radical, and then multiply in front of the radical. An example is:

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The multiplication of a radical is easier.

Learn more about polynomial on:

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3 0
2 years ago
A right triangle has a hypotenuse length of 12, and one side length of 9. find the missing side length. do the side lengths form
Rina8888 [55]

Use the Pythagorean theorem:

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A Pythagorean triple consists of three positive integers a, b and c, such that

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Answer:

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