Question

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

Answer 1

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

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

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=(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=(1)/(8) \implies y&=4\left( (1)/(8)\right)^3\\y&=4 \cdot (1)/(512)\\y&=(1)/(128)\end{aligned}`

Therefore, the solutions are:

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

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

Answer 2

• (0, 0)
• (1/8, 1/128)

=====================

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