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kaheart [24]
3 years ago
5

What is the solution of the linear-quadratic system of equations? y=x' + 5x + 7 y = x +4

Mathematics
1 answer:
grin007 [14]3 years ago
4 0

Answer:

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=x^2+5x+7,\:y=x+4\mathrm{\:are\:}

\begin{pmatrix}x=-1,\:&y=3\\ x=-3,\:&y=1\end{pmatrix}

Step-by-step explanation:

Given the system of the equations

y=x^2\:+\:5x\:+\:7

y\:=\:x\:+4

steps to solve the system of the equations

\begin{bmatrix}y=x^2+5x+7\\ y=x+4\end{bmatrix}

subtract the equations

y=x^2+5x+7

-

\underline{y=x+4}

y-y=x^2+5x+7-\left(x+4\right)

0=x^2+4x+3

solving

0=x^2+4x+3

\left(x+1\right)\left(x+3\right)=0

x+1=0\quad \mathrm{or}\quad \:x+3=0

x=-1,\:x=-3

\mathrm{Plug\:the\:solutions\:}x=-1,\:x=-3\mathrm{\:into\:}y=x^2+5x+7

substitute x=-1 into y=x^2+5x+7

y=x^2+5x+7

y=\left(-1\right)^2-5\cdot \:1+7

y=1-5+7

y=3

Now substitute x=-3 into y=x^2+5x+7

y=x^2+5x+7

y=\left(-3\right)^2+5\left(-3\right)+7

y=3^2-15+7

y=3^2-8

y=9-8

y=1

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=x^2+5x+7,\:y=x+4\mathrm{\:are\:}

\begin{pmatrix}x=-1,\:&y=3\\ x=-3,\:&y=1\end{pmatrix}

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It's a

Step-by-step explanation:

This is a quadratic expression with the highest power of X being 2.

The coefficient of X² = 1

Note that the general form of a quadratic equation is aX² + bX + C

In this question,

a = 1

b = 7

C = 10

First step is to multiply a and C ( a x C)

= 1 X 10 = 10

Then, find 2 factors of 10 which can both add up to give 7, the coefficient of X given in the question.

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