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3 years ago

In order to solve the following system of equations by subtraction, which of

Mathematics

1 answer:

uysha [10]3 years ago

7 0

Answer: I think it's C.Multiply the top equation by 3 and the bottom equation by 4

Step-by-step explanation:

Multiplying the top equation by 3 you get:

3(4x – 2y) = 3∗7

12x - 6y = 21 (eq.1)

Multiplying the bottom equation by 4 you get:

4(3x – 3y) = 4∗15

12x - 12y = 60 (eq.2)

Now, if you subtact eq.2 from eq.1 you get:

(12x - 6y = 21 )

(12x - 12y = 60)

___________________

0x + [- 6y -(-12y)] = 21 - 60

Carrying out the calculation you get:

6y = -39

y = -39/6

y = -13/2

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mrs_skeptik [129]

Answer:

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Step-by-step explanation:

Given that -1 is a zero of multiplicity of two for the polynomial $h(x) = x^4+23x^2+50x+26.$ This means that $x-(-1)=x+1$ is twice the linear factor.

Rewrite the polynomial $h(x) = x^4+23x^2+50x+26$ as follows:

$h(x)\\ \\= x^4+23x^2+50x+26\\ \\=x^4+x^3-x^3-x^2+24x^2+24x+26x+26\\ \\=x^3(x+1)-x^2(x+1)+24x(x+1)+26(x+1)\\ \\=(x+1)(x^3-x^2+24x+26)\\ \\=(x+1)(x^3+x^2-2x^2-2x+26x+26)\\ \\=(x+1)(x^2(x+1)-2x(x+1)+26(x+1))\\ \\=(x+1)(x+1)(x^2-2x+26)$

Find linear factors of the quadratic polynomial $x^2-2x+26$ :

$D=(-2)^2-4\cdot 1\cdot 26=4-104=-100=100i^2\\ \\\sqrt{D}=\sqrt{100i^2}=10i\\ \\x_{1,2}=\dfrac{-(-2)\pm 10i}{2\cdot 1}=\dfrac{2\pm 10i}{2}=1\pm 5i$

Hence,

$x^2-2x+26=(x-(1+5i))(x-(1-5i))=(x-1-5i)(x-1+5i)$

and the initial polynomial factorization is

$h(x)=(x+1)^2(x-1-5i)(x-1+5i)$

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