Question

solve the following system of equations algebraically:
$y = 4x^2 + 19x - 6$
$y = 2x - 10$
answer attempt 1 out of 2
two solutions
and

Explanation:

Step1: Set the equations equal

Since both equal \( y \), set \( 4x^2 + 19x - 6 = 2x - 10 \).

Step2: Simplify the equation

Subtract \( 2x \) and add \( 10 \): \( 4x^2 + 17x + 4 = 0 \).

Step3: Factor the quadratic

Find two numbers: \( 16 \) and \( 1 \), so \( 4x^2 + 16x + x + 4 = 0 \). Factor: \( 4x(x + 4) + 1(x + 4) = 0 \), then \( (4x + 1)(x + 4) = 0 \).

Step4: Solve for \( x \)

Set each factor to zero: \( 4x + 1 = 0 \Rightarrow x = -\frac{1}{4} \); \( x + 4 = 0 \Rightarrow x = -4 \).

Answer:

The solutions for \( x \) are \( -\frac{1}{4} \) and \( -4 \). (To find \( y \), substitute back: for \( x = -\frac{1}{4} \), \( y = 2(-\frac{1}{4}) - 10 = -\frac{21}{2} \); for \( x = -4 \), \( y = 2(-4) - 10 = -18 \). So the solution pairs are \( (-\frac{1}{4}, -\frac{21}{2}) \) and \( (-4, -18) \). If only \( x \) is needed, the answer is \( -\frac{1}{4} \) and \( -4 \).)