Question

Question 5 of 15
What would be the best first step in solving this system
x²-3x + 2y = -4
y = 3x + 2

Answer 1

Isolate y on both equations then equate

• 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

Answer 2

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 second equation into the first equation, 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 equals zero:

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

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

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

Now use the Quadratic Formula to solve for x.

Quadratic Formula

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

Define the variables:

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

Substitute the values into the formula 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:

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