Question

What are the solutions of the system?

y = x^2-3x+2
y = 4x-4

A. (1, 0) and (–6, –28)

B. (1, 20) and (6, 0)

C. (1, 0) and (6, 20)

D. no solution

A, B, C, or D? Please explain :)

Answer 1

solve
\[ x^2-3x+2 = 4x-4\]

Answer 2

since y is expressed in two different ways you can set them equal to each other to find one or more x values that satisfies the equation

x^2-3x+2 = 4x-4
x^2-3x+2-4x-4=0
x^2-7x-2 = 0

It does not seem to be easily factored so you will need the quadratic equation to solve this. Since it will likely come out to a decimal, the solution is no solution.

Answer 3

Thank you @seitys. Are you positive that there is not a solution?

Answer 4

there is are x values that satisfy both equations. you have to plug it into the quadratic equation to get the answer. Usually with small values <10, if it is not factor-able, it comes out to a decimal number which is not represented in your solution set.

Answer 5

Okay then, thank you for your input and answer.