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3 years ago

1/3x+2=3/4 how would I go about solving this???

Mathematics

1 answer:

Marizza181 [45]3 years ago

8 0

Answer combine 1/2 and x.

X/2+2/3=1/6

Subtract 2/3 from both sides of the equation

X/2=1/6-2/3

Simplify the right side of the equation

-2/3 as a fraction with a common denominator, multiply by 2/2

X/2=1/6-2/3 x 2/2

Multiply 3 by 2

X/2 = 1/6 - 2/6 x 2

Combine the numerators over the common denominator

X/2 = 1-2 x2/ 6

Simplify the numerator

Multiply by -2 by 2

X/2 =1-4/6

Subtract 4 from 1

X/2= -3/6

Cancel the common factor of -3 and 6

Factor 3 out of -3

X/2 = 3(-1)/6

Cancel the common factor of 6

X/2 = 3 .-1/3.2

Rewrite the expression

X/2 = -1/2

Move the negative in front of the fraction

X/2= -1/2

Since both sides have the same Denominator the numerators must be equal.

X=-1

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GuDViN [60]

Answer:

vertex = (0, -4)

equation of the parabola: $y=3x^2-4$

Step-by-step explanation:

Given:

y-intercept of parabola: -4
parabola passes through points: (-2, 8) and (1, -1)

Vertex form of a parabola: $y=a(x-h)^2+k$

(where (h, k) is the vertex and $a$ is some constant)

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$\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}$

Substitute point (-2, 8) and $ah^2+k=-4$ into the equation:

$\begin{aligned}\textsf{At}\:(-2,8) \implies a(-2-h)^2+k &=8\\a(4+4h+h^2)+k &=8\\4a+4ah+ah^2+k &=8\\\implies 4a+4ah-4&=8\\4a(1+h)&=12\\a(1+h)&=3\end{aligned}$

Substitute point (1, -1) and $ah^2+k=-4$ into the equation:

$\begin{aligned}\textsf{At}\:(1.-1) \implies a(1-h)^2+k &=-1\\a(1-2h+h^2)+k &=-1\\a-2ah+ah^2+k &=-1\\\implies a-2ah-4&=-1\\a(1-2h)&=3\end{aligned}$

Equate to find h:

$\begin{aligned}\implies a(1+h) &=a(1-2h)\\1+h &=1-2h\\3h &=0\\h &=0\end{aligned}$

Substitute found value of h into one of the equations to find a:

$\begin{aligned}\implies a(1+0) &=3\\a &=3\end{aligned}$

Substitute found values of h and a to find k:

$\begin{aligned}\implies ah^2+k&=-4\\(3)(0)^2+k &=-4\\k &=-4\end{aligned}$

Therefore, the equation of the parabola in vertex form is:

$\implies y=3(x-0)^2-4=3x^2-4$

So the vertex of the parabola is (0, -4)

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