Algebra 1, sem.; show work! what are the solutions to the system of equations? y=4x2+6x−3 y=2x−3

Both of the equations above are equal to y. We can use transitivity (a is the same as b and a is the same as c, so b and c are the same) and say that they are equal to one another. Call the equations A and B to keep them straight.

A) y = 4x² + 6x − 3

B) y = 2x − 3

2x - 3 = 4x² + 6x − 3 put equation B into A, by transitivity (or substitution)

2x = 4x² + 6x Add 3 on both sides - notice it goes away on both sides.

0 = 4x² + 4x subtract 2x on both sides.

We set the second degree equation equal to zero and use the Zero Product Property - if a product multiplies to zero, so do its pieces - and factoring. Here we use group factoring: 4x is in common with both and so we can group it and pull it out.

0 = 4x² + 4x

0 = 4x (x + 1) Factor the 4x

0 = 4x OR 0 = x + 1 Apply the Zero Product Property

Now we solve both equations for x.

0 = 4x so x 0

0 = x + 1 so x = -1

Next, we take the two solutions to find points for y. We substitute them into equation B

When x = 0, y = 2(0) - 3. Then y = (2 * 0) - 3 = -3. This makes (0, -3) a solution.

When x = -1 y = 2(-1) - 3. Then y = (2 * -1) -3 = -2 - 3 = -5. This makes (-1, -5) a solution.

Thus, (0, -3) and (-1, -5) are solutions.

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