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pishuonlain [190]
2 years ago
8

Question 5 of 15

Mathematics
2 answers:
Olenka [21]2 years ago
7 0

Answer:

Substitute y = 3x + 2 into x² - 3x + 2y = -4

Step-by-step explanation:

Given system of equations:

\begin{cases}x^2-3x+2y=-4\\y=3x+2\end{cases}

Since y is already isolated in the second equation, the best first step would be to substitute the <u>second equation</u> into the <u>first equation</u>, then solve for x.

Substitute equation 2 into equation 1:

\implies x^2-3x+2(3x+2)=-4

Expand the brackets and simplify so that the equation <u>equals zero</u>:

\implies x^2-3x+6x+4=-4

\implies x^2+3x+4+4=0

\implies x^2+3x+8=0

Now use the Quadratic Formula to solve for x.

<u>Quadratic Formula</u>

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

<u>Define</u> the variables:

\implies a=1, \quad b=3, \quad c=8

Substitute the values into the <u>formula</u> and solve for x:

\implies x=\dfrac{-3 \pm \sqrt{3^2-4(1)(8)}}{2(1)}

\implies x=\dfrac{-3 \pm \sqrt{-23}}{2}

\implies x=\dfrac{-3 \pm \sqrt{-1 \cdot 23}}{2}

\implies x=\dfrac{-3 \pm \sqrt{-1}\sqrt{23}}{2}

Remember that i^2=-1

\implies x=\dfrac{-3 \pm i\sqrt{23}}{2}

Therefore, the solutions to the system of equations are:

x=\dfrac{-3 + i\sqrt{23}}{2}, \quad \dfrac{-3 - i\sqrt{23}}{2}

Learn more about systems of equations here:

brainly.com/question/27520807

Licemer1 [7]2 years ago
6 0
<h3>Isolate y on both equations then equate </h3>

  • x²-3x+2y=-4
  • 2y=-x²+3x-4
  • y=-x²/2+3/2x-2

Now you can equate it with second one to get x

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