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2 years ago

Explain how you would create opposite terms for the x-variable in this system. 3x-6y=-8, -1/4x+4y=6

Mathematics

2 answers:

Molodets [167]2 years ago

6 0

Your answer would be, You can use the Elimination Method, by Multiplying, One or Both Equations by a Constant.

Solve System of Linear Equation:

3x - 6y = -8, -1/4x + 4y = 6 ====> y = 32 /21; x = 8 / 21

Steps:

[ 3x - 6y = -8, -1 / 4x + 4y = 6 ]

Isolate x for 3x - 6y = -8: x = -8 + 6y / 3

3x - 6y = -8

Add 6y both sides:

3x - 6y + 6y = -8 + 6y

Simplify:

3x = - 8 + 6y

Divide Both sides by 3:

3x / x = -8 / 3 + 6y / 3

Simplify:

x = -8 + 6y / 3

Substitute:

x = -8 + 6y / 3

[ 1 / 4 * -8 + 6y / 3 + 4y = 6 ]

Isolate:

-1 / 4 -8 + 6y / 3 + 4y = 6

y = 32 / 21

For x = -8 + 6y / 3

Substitute:

y = 32 / 21

x = -8 + 6 / 3 * 32 / 21

-8 + 6 / 3 * 32 / 21

x = 8 / 21

The Solution to the Systems of Equations are: y = 32 / 21; x = 8 / 21

Hope that helps!!! : )

suter [353]2 years ago

4 0

To create opposite terms for the x-variable, you need to multiply the second equation by a number that will result in the x term becoming –3x. That number would need to be a multiple of the denominator, which is 3. Multiplying the second equation by 9 would create opposite terms for the x-variable.

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Answer:

a) $P (x) = (x + 3) (x-1) (x-4)$

b) $P (x) = (2x + 5) (5x - 4) (x-6)$

c) $P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

Step-by-step explanation:

For the question a * you need to find a polynomial of degree 3 with zeros in -3, 1 and 4.

This means that the polynomial P(x) must be zero when x = -3, x = 1 and x = 4.

Then write the polynomial in factored form.

$P (x) = (x + 3) (x-1) (x-4)$

Note that this polynomial has degree 3 and is zero at x = -3, x = 1 and x = 4.

For question b, do the same procedure.

Degree: 3

Zeros: -5/2, 4/5, 6.

The factors are

$x = -\frac{5}{2}\\\\x +\frac{5}{2} = 0\\\\(2x +5) = 0$

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$x =\frac{4}{5}\\\\x-\frac{4}{5} = 0\\\\(5x-4) = 0$

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

$x = 6\\\\(x-6) = 0$

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$P (x) = (2x + 5) (5x - 4) (x-6)$

Finally for the question c we have

Degree: 5

Zeros: -3, 1, 4, -1

Multiplicity 2 in -1

$x = -3\\\\(x-3) = 0$

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$x = 1\\\\(x-1) = 0$

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$x = 4\\\\(x-4) = 0$

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$x = -1\\\\(x + 1) = 0$

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$P (x) = (x-3) (x-1) (x-4) (x + 1) ^ 2$

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