What is the relationship between the lines determined by the following two equations?

Mathematics

1 answer:

oksano4ka [1.4K]3 years ago

4 0

Answer:

C. They are the same line.

Step-by-step explanation:

In order to compare the linear equations given, they need to be in the same form. The best form in order to evaluate slope and y-intercept is slope-intercept form, y = mx + b. Since the second equation is already in slope-intercept form, we need to use inverse operations to convert the first equation:

6x - 2y = 16 ---- 6x - 2y - 6x = 16 - 6x ---- -2y = -6x + 16

-2y/-2 = -6x/-2 + 16/-2

y = 3x - 8

Since both equations are in the form y = 3x - 8, then they are both the same line.

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The given question is a quadratic equation and we can use several methods to get the solutions to this question. The solution to the equation are 3/4 and -5/6 and the greater of the two solutions is 3/4

<h3>Quadratic Equation</h3>

Quadratic equation are polynomials with a second degree as it's highest power.

An example of a quadratic equation is

$y = ax^2 + bx + c$

The given quadratic equation is $24x^2 + 2x = 15$

Let's rearrange the equation

$24x^2 + 2x = 15\\24x^2 + 2x - 15 = 0$

This implies that

a = 24
b = 2
c = -15

The equation or formula of quadratic formula is given as

$y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}$

We can substitute the values into the equation and solve

$y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}\\y = \frac{-2 +- \sqrt{2^2 -4 * 24 * (-15)} }{2*24} \\y = \frac{-2+-\sqrt{4+1440} }{48} \\y = \frac{-2+-\sqrt{1444} }{48} \\y = \frac{-2+- 38}{48} \\y = \frac{-2+38}{48} \\y = \frac{3}{4}\\ \\or\\y = \frac{-2-38}{48} \\y = \frac{-40}{48} \\y = -\frac{5}{6}$