Question

NO LINKS!! Use the method of substitution to solve the system. (if there's no solution, enter no solution). Part 4z​

Answer 1

Answer:

• (0, 0)
• (1/8, 1/128)

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Given system

• 2y = x²
• y = 4x³

Double the second equation

• 2y = x²
• 2y = 8x³

Solve by elimination

• x² = 8x³
• x²(8x - 1) = 0
• x = 0 and 8x - 1 = 0
• x = 0 and x = 1/8

Find the value of y

• x = 0 ⇒ y = 0
• x = 1/8 ⇒ y = 4*(1/8)³ = 1/128

Answer 2

Answer:

$(x,y)=\left(\; \boxed{0,0} \; \right)\quad \textsf{(smaller x -value)}$

$(x,y)=\left(\; \boxed{\dfrac{1}{8}, \dfrac{1}{128}} \; \right)\quad \textsf{(larger x -value)}$

Step-by-step explanation:

Given system of equations:

$\begin{cases}2y=x^2\\\;\;y=4x^3\end{cases}$

To solve by the method of substitution, substitute the second equation into the first equation, rearrange so that the equation equals zero, then factor:

$\begin{aligned}y=4x^3 \implies 2(4x^3)&=x^2\\8x^3&=x^2\\8x^3-x^2&=0\\x^2(8x-1)&=0\end{aligned}$

Apply the zero-product property to solve for x:

$\implies x^2=0 \implies x=0$

$\implies 8x-1=0 \implies x=\dfrac{1}{8}$

Substitute the found values of x into the second equation and solve for y:

$\begin{aligned}x=0 \implies y&=4(0)^3\\y&=0\end{aligned}$

$\begin{aligned}x=\dfrac{1}{8} \implies y&=4\left( \dfrac{1}{8}\right)^3\\y&=4 \cdot \dfrac{1}{512}\\y&=\dfrac{1}{128}\end{aligned}$

Therefore, the solutions are:

$(x,y)=\left(\; \boxed{0,0} \; \right)\quad \textsf{(smaller x -value)}$

$(x,y)=\left(\; \boxed{\dfrac{1}{8}, \dfrac{1}{128}} \; \right)\quad \textsf{(larger x -value)}$