Question

LOT OF POINTS

Answer 1

Answer:

Given equations: y = 3x² + x - 6; y = x - 4

So, Plug the value of value of y in other equation

• 3x² + x - 6 = x - 4
• 3x² + x - x - 6 + 4 = 0
• 3x² - 2 = 0
• 3x² = 2
• x² = 2/3
• x = √2/3

Answer 2

Answer:

$\displaystyle ( x_{1},y_{1}) = (0.82, - 3.18) \\ ( x_{2},y_{2}) = ( - 0.82, - 4.82)$

Step-by-step explanation:

we are given a system of quadratic and linear equation

we want to figure out x and y

in other words the coordinates where the linear function intercept quadrilateral function

to do so

you can use substitution method

since y equals to both equation so substitute:

$\displaystyle {3x}^{2} + x - 6 = x - 4$

move right hand side expression to left hand side and change its sign so there's only 0 left on the left hand side:

$\displaystyle {3x}^{2} + x - 6 - x + 4= 0$

simplify addition:

$\displaystyle {3x}^{2} -2=0$

add 2 to both sides:

$\displaystyle {3x}^{2} = 2$

divide both sides by 3

$\displaystyle \frac{ {3x}^{2} }{3} = \frac{2}{3}$

$\displaystyle {x}^{2} = \frac{2}{3}$

square root both sides:

$\displaystyle {x} = \sqrt{\frac{2}{3} } \\ x = \frac{ \sqrt{2} }{ \sqrt{3} }$

rationalise the denominator by multiplying √3/√3:

$\displaystyle x = \pm\frac{ \sqrt{6} }{3} = \pm0.82$

now let's figure out y

substitute the value of x to the linear equation:

$y = \pm0.82 - 4$

when positive

$y = - 3.18$

when negative

$y = - 4.82$