The Space between Ti.IVIe

If we use substitution, then

$7x+49=19x^{2}+7x+30$

When we set one side of the equation equal to 0, we get:

$0=19x^{2}-19$

When we divide both sides by 19, we get:

$0=x^{2}-1$

which (as a universal and essential fact can be factored into:

$0=(x+1)\times(x-1)$

The only 2 solutions are $x=\pm1$ .

Now, since you think smarter, not harder, you look at the answer choices to find the only answer that contains something along the lines of " $(1, y_1)~\text{and}~(-1,y_2)$ ," and select that answer.

If a system of equations includes a linear equation and a quadratic equation (parabola, circle, ellipse, or hyperbola), then solve using substitution.

Look at the system of equations: y=19x^2+7x+30 Parabola y=7x+49 Line

Since there is a linear equation, it's best to use substitution.

Notice that the parabolic equation has an x2 term, but no y2 term. To avoid squaring, it's best to isolate y in the linear equation. Luckily, it's already isolated: y=7x+49.

Substitute y=7x+49 into the parabolic equation and solve for x.

y = 19x^2+7x+30 7x+49 = 19x^2+7x+30 19 = 19x^62 1 = x^2 ±1 = x

Finally, plug x=1 and x= -1 into one of the original equations to find the solutions. It's easier to use the linear equation.

Plug in x=1. y = 7x+49 = 7(1)+49 = 7+49 = 56 The solution is (1,56). Plug in x= -1. y = 7x+49 = 7(-1)+49= -7+49 = 42 The solution is (-1,42).