Question

NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 8z​

Answer 1

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

`(x,y)=\left(\; \boxed{(1)/(10),(1)/(200)} \; \right)\quad \textsf{(larger $x$-value)}`

Explanation:

Given system of equations:

`\begin{cases}2y=x^2\\ \;\;y=5x^3\end{cases}`

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

`\begin{aligned}y=5x^3 \implies 2(5x^3)&=x^2\\10x^3&=x^2\\10x^3-x^2&=0\\\end{aligned}`

Factor the equation:

`\begin{aligned}10x^3-x^2&=0\\x^2(10x-1)&=0\end{aligned}`

Apply the zero-product property and solve for x:

`x^2=0 \implies x=0`

`10x-1=0 \implies x=(1)/(10)`

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

`\begin{aligned}x=0 \implies y&=5(0)^3\\y&=0\end{aligned}`

`\begin{aligned}x=0 \implies y&=5\left((1)/(10)\right)^3\\y&=5 \cdot (1)/(1000)\\y&=(1)/(200)\end{aligned}`

Therefore, the solutions are:

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

`(x,y)=\left(\; \boxed{(1)/(10),(1)/(200)} \; \right)\quad \textsf{(larger $x$-value)}`

Answer 2

• (0, 0)
• (0.1, 0.005)

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Given system

• 2y = x²
• y = 5x³

Substitute the value of y into first equation

• 2*5x³ = x²
• 10x³ - x² = 0
• x²(10x - 1) = 0
• x = 0 and 10x - 1 = 0
• x = 0 and x = 0.1

Find the value of y

• x = 0 ⇒ y = 5*0³ = 0
• x = 0.1 ⇒ y = 5*(0.1)³ = 0.005