Find the Points of Intersection y=7x-3 y=-2x^2+1

$y = 7 x - 3$ $y = - 2 x^{2} + 1$

Step 1

Eliminate the equal sides of each equation and combine.

$7 x - 3 = - 2 x^{2} + 1$

Step 2

Solve $7 x - 3 = - 2 x^{2} + 1$ for $x$ .

Add $2 x^{2}$ to both sides of the equation.

$7 x - 3 + 2 x^{2} = 1$

Subtract $1$ from both sides of the equation.

$7 x - 3 + 2 x^{2} - 1 = 0$

Subtract $1$ from $- 3$ .

$7 x + 2 x^{2} - 4 = 0$

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$7 u + 2 u^{2} - 4 = 0$

Factor by grouping.

Reorder terms.

$2 u^{2} + 7 u - 4 = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 4 = - 8$ and whose sum is $b = 7$ .

Factor $7$ out of $7 u$ .

$2 u^{2} + 7 (u) - 4 = 0$

Rewrite $7$ as $- 1$ plus $8$

$2 u^{2} + (- 1 + 8) u - 4 = 0$

Apply the distributive property.

$2 u^{2} - 1 u + 8 u - 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 u^{2} - 1 u) + 8 u - 4 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (2 u - 1) + 4 (2 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

$(2 u - 1) (u + 4) = 0$

Replace all occurrences of $u$ with $x$ .

$(2 x - 1) (x + 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 x - 1 = 0$

$x + 4 = 0 + y = - 2 x^{2} + 1$

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(2 x - 1) (x + 4) = 0$ true.

$x = \frac{1}{2}, - 4$

Step 3

Evaluate $y$ when $x = \frac{1}{2}$ .

Substitute $\frac{1}{2}$ for $x$ .

$y = - 2 {(\frac{1}{2})}^{2} + 1$

Simplify $- 2 {(\frac{1}{2})}^{2} + 1$ .

Simplify each term.

Apply the product rule to $\frac{1}{2}$ .

$y = - 2 \frac{1^{2}}{2^{2}} + 1$

One to any power is one.

$y = - 2 \frac{1}{2^{2}} + 1$

Raise $2$ to the power of $2$ .

$y = - 2 (\frac{1}{4}) + 1$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$y = 2 (- 1) \frac{1}{4} + 1$

Factor $2$ out of $4$ .

$y = 2 \cdot - 1 \frac{1}{2 \cdot 2} + 1$

Cancel the common factor.

$y = 2 \cdot - 1 \frac{1}{2 \cdot 2} + 1$

Rewrite the expression.

$y = - 1 (\frac{1}{2}) + 1$

Rewrite $- 1 (\frac{1}{2})$ as $- (\frac{1}{2})$ .

$y = - \frac{1}{2} + 1$

Simplify the expression.

Write $1$ as a fraction with a common denominator.

$y = - \frac{1}{2} + \frac{2}{2}$

Combine the numerators over the common denominator.

$y = \frac{- 1 + 2}{2}$

Add $- 1$ and $2$ .

$y = \frac{1}{2}$

Step 4

Evaluate $y$ when $x = - 4$ .

Substitute $- 4$ for $x$ .

$y = - 2 {(- 4)}^{2} + 1$

Simplify $- 2 {(- 4)}^{2} + 1$ .

Simplify each term.

Raise $- 4$ to the power of $2$ .

$y = - 2 \cdot 16 + 1$

Multiply $- 2$ by $16$ .

$y = - 32 + 1$

Add $- 32$ and $1$ .

$y = - 31$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(\frac{1}{2}, \frac{1}{2})$

$(- 4, - 31)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{1}{2}, \frac{1}{2}), (- 4, - 31)$

Equation Form:

$x = \frac{1}{2}, y = \frac{1}{2}$

$x = - 4, y = - 31$

Step 7

Find the Points of Intersection y=7x-3 y=-2x^2+1

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