Question

question #9
which of the following are a solution to the quadratic - linear system of equations? -x² + 2y - 3 = 0 -3x + y + 1 = 0
i. (5,14)
ii. (1, 2)
iii. (-5, -16)
iv. (-1 - 4)
i and ii
iii only
iii and iv
ii only

Explanation:

Step1: Rearrange the linear equation

From $-3x + y+1 = 0$, we can express $y$ in terms of $x$ as $y=3x - 1$.

Step2: Substitute $y$ into the quadratic equation

Substitute $y = 3x - 1$ into $-x^{2}+2y - 3=0$. We get $-x^{2}+2(3x - 1)-3 = 0$. Expand it: $-x^{2}+6x-2 - 3=0$, which simplifies to $-x^{2}+6x - 5=0$. Multiply through by - 1 to get $x^{2}-6x + 5=0$.

Step3: Factor the quadratic equation

Factor $x^{2}-6x + 5=0$ as $(x - 1)(x - 5)=0$.

Step4: Solve for $x$

Set each factor equal to zero: $x-1=0$ gives $x = 1$; $x - 5=0$ gives $x = 5$.

Step5: Find corresponding $y$ - values

When $x = 1$, $y=3\times1-1=2$ (using $y = 3x - 1$). When $x = 5$, $y=3\times5-1=14$.

Answer:

A. I and II