Question

1. solve the system by elimination.

$-5x + 29 = y - x^2$
$x^2 + y = 2x^2 - 1$

a. $square$ $(5, 28)$
b. $square$ $(6, 35)$
c. $square$ $(7, 10)$
d. $square$ $(9, 22)$
e. $square$ $(10, 31)$
f. $square$ no real solution

Explanation:

Step1: Rewrite the system of equations

We have the system:

$$
LATEXBLOCK0
$$

From the second equation, we can solve for \(y\): \(y=2x^2 - 1 - x^2=x^2 - 1\).

Step2: Substitute \(y\) into the first equation

Substitute \(y = x^2 - 1\) into the first equation \(-5x + 29 = y - x^2\):

$$ -5x + 29=(x^2 - 1)-x^2 $$

Simplify the right - hand side: \((x^2 - 1)-x^2=-1\). So the equation becomes \(-5x + 29=-1\).

Step3: Solve for \(x\)

Subtract 29 from both sides: \(-5x=-1 - 29=-30\).
Divide both sides by \(-5\): \(x = \frac{-30}{-5}=6\).

Step4: Solve for \(y\)

Substitute \(x = 6\) into \(y=x^2 - 1\). Then \(y = 6^2-1=36 - 1 = 35\).

Answer:

b. (6, 35)