Question

Please help me with this

Answer 1

C

Explanation:

Given the 2 equations

y = 2x - 3 → (1)

y = x² - 3 → (2)

Substitute y = x² - 3 into (1)

x² - 3 = 2x - 3 ( subtract 2x - 3 from both sides )

x² - 2x = 0 ← factor out x from each term

x(x - 2) = 0

Equate each factor to zero and solve for x

x = 0

x - 2 = 0 ⇒ x = 2

Substitute these values into (1) for corresponding values of y

x = 0 : y = 2(0) - 3 = 0 - 3 = - 3 ⇒ (0, - 3 )

x = 2 : y = 2(2) - 3 = 4 - 3 = 1 ⇒ (2, 1 )

Answer 2

Option c

Explanation:

A system of equations are given to us. And we need to solve them . The given system is

`\begin{cases} y = 2x - 3\dots 1 \\ y = x^2 - 3\dots 2 \end{cases}`

We numbered the equations here . Now put the value of equation 1 in equation 2 that is substituting y = 2x - 3 in eq. 2 .

`\implies y = x^2 - 3 \\\\\implies 2x - 3 = x^2 - 3 \\\\\implies x^2 - 2x = 0 \\\\\implies x(x-2) = 0 \\\\\implies\red{ x = 0 , 2 }`

We got two values of x as 0 & 2 . Alternatively substituting these values we have ,

`\implies y = 2 x - 3 \\\\\implies y = 2(0)-3 \qquad or \qquad y = 2(2)-3 \\\\\implies y = 0-3 \qquad or \qquad 4 - 3 \\\\\implies \red{ y = -3 , 1 }`

Thefore the required answer is ,

`\red{Option\:c} \begin{cases} (0,-3) \\ (2,1) \end{cases}`