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3 years ago

Need help again ...these questions are hard

Mathematics

2 answers:

kherson [118]3 years ago

6 0

The answer is
(0, 5)
(1, 3)

OLga [1]3 years ago

4 0

Answer:

Number 1 and 4

Step-by-step explanation:

To do this substitute the different values of x and y into the equations and see if they come out true or not.

1. (4, -3)

Put x = 4 and y = -3 into both equations:

2x + y = 5 --> 2(4) - 3 = 5 --> 8 - 3 = 5 --> 5=5 (true)

6x + 3y = 15 --> 6(4) + 3(-3) = 15 --> 24 - 9 = 15 --> 15 = 15 (true

So number 1 works

2. (2, 4)

Put x = 2 and y = 4 into both equations:

2x + y = 5 --> 2(2) + 4 = 5 --> 4 + 4 = 5 --> 8 = 5 (NOT true)

Number 2 does not work

3. (1, 3)

Put x = 1 and y = 3 into both equations:

2x + y = 5 --> 2(1) + 3 = 5 --> 2+ 3 = 5 --> 5 = 5 (true)

6x + 3y = 15 --> 6(1) + 3(9) = 15 --> 6 + 27 = 15 --> 33 = 15 (not true)

Number 3 does not work

4. (0, 5)

Put x = 0 and y = 5 into both equations:

2x + y = 5 --> 2(0) + 5 = 5 --> 5 = 5 (true)

6x + 3y = 15 --> 6(0) + 3(5) = 15 --> 15 = 15 (true)

Number 4 does work

Number 1 and number 4 are the answers.

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What are the solutions of the equation 9x4 – 2x2 – 7 = 0? Use u substitution to solve. tions of the equation 9

skelet666 [1.2K]

Answer:

Step-by-step explanation:

Let $u^2=x^4\\u = x^2$

Subbing in:

$9u^2-2u-7=0$

a = 9, b = -2, c = -7

The product of a and c is the aboslute value of -63, so a*c = 63. We need 2 factors of 63 that will add to give us -2. The factors of 63 are {1, 63}, (3, 21}, {7, 9}. It looks like the combination of -9 and +7 will work because -9 + 7 = -2. Plug in accordingly:

$9u^2-9u+7u-7=0$

Group together in groups of 2:

$(9u^2-9u)+(7u-7)=0$

Now factor out what's common within each set of parenthesis:

$9u(u-1)+7(u-1)=0$

We know this combination "works" because the terms inside the parenthesis are identical. We can now factor those out and what's left goes together in another set of parenthesis:

$(u-1)(9u+7)=0$

Remember that $u=x^2$

so we sub back in and continue to factor. This was originally a fourth degree polynomial; that means we have 4 solutions.

$(x^2-1)(9x^2+7)=0$

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis. Factoring $(x^2-1)$ gives us that x = 1 and -1. The other set is a bit more tricky. If