Solve the System of Equations y=2x^2-5x+5 y-x=5

$y = 2 x^{2} - 5 x + 5$ $y - x = 5$

Step 1

Replace all occurrences of $y$ with $2 x^{2} - 5 x + 5$ in each equation.

Replace all occurrences of $y$ in $y - x = 5$ with $2 x^{2} - 5 x + 5$ .

$(2 x^{2} - 5 x + 5) - x = 5$

$y = 2 x^{2} - 5 x + 5$

Simplify the left side.

Subtract $x$ from $- 5 x$ .

$2 x^{2} - 6 x + 5 = 5$

$y = 2 x^{2} - 5 x + 5$

$2 x^{2} - 6 x + 5 = 5$

$y = 2 x^{2} - 5 x + 5$

$2 x^{2} - 6 x + 5 = 5$

$y = 2 x^{2} - 5 x + 5$

Step 2

Solve for $x$ in $2 x^{2} - 6 x + 5 = 5$ .

Subtract $5$ from both sides of the equation.

$2 x^{2} - 6 x + 5 - 5 = 0$

$y = 2 x^{2} - 5 x + 5$

Combine the opposite terms in $2 x^{2} - 6 x + 5 - 5$ .

Subtract $5$ from $5$ .

$2 x^{2} - 6 x + 0 = 0$

$y = 2 x^{2} - 5 x + 5$

Add $2 x^{2} - 6 x$ and $0$ .

$2 x^{2} - 6 x = 0$

$y = 2 x^{2} - 5 x + 5$

$2 x^{2} - 6 x = 0$

$y = 2 x^{2} - 5 x + 5$

Factor $2 x$ out of $2 x^{2} - 6 x$ .

Factor $2 x$ out of $2 x^{2}$ .

$2 x (x) - 6 x = 0$

$y = 2 x^{2} - 5 x + 5$

Factor $2 x$ out of $- 6 x$ .

$2 x (x) + 2 x (- 3) = 0$

$y = 2 x^{2} - 5 x + 5$

Factor $2 x$ out of $2 x (x) + 2 x (- 3)$ .

$2 x (x - 3) = 0$

$y = 2 x^{2} - 5 x + 5$

$2 x (x - 3) = 0$

$y = 2 x^{2} - 5 x + 5$

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

$x = 0$

$x - 3 = 0$

$y = 2 x^{2} - 5 x + 5$

Set $x$ equal to $0$ .

$x = 0$

$y = 2 x^{2} - 5 x + 5$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

$y = 2 x^{2} - 5 x + 5$

Add $3$ to both sides of the equation.

$x = 3$

$y = 2 x^{2} - 5 x + 5$

$x = 3$

$y = 2 x^{2} - 5 x + 5$

The final solution is all the values that make $2 x (x - 3) = 0$ true.

$x = 0, 3$

$y = 2 x^{2} - 5 x + 5$

$x = 0, 3$

$y = 2 x^{2} - 5 x + 5$

Step 3

Replace all occurrences of $x$ with $0$ in each equation.

Replace all occurrences of $x$ in $y = 2 x^{2} - 5 x + 5$ with $0$ .

$y = 2 {(0)}^{2} - 5 \cdot 0 + 5$

$x = 0$

Simplify the right side.

Simplify $2 {(0)}^{2} - 5 \cdot 0 + 5$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 2 \cdot 0 - 5 \cdot 0 + 5$

$x = 0$

Multiply $2$ by $0$ .

$y = 0 - 5 \cdot 0 + 5$

$x = 0$

Multiply $- 5$ by $0$ .

$y = 0 + 0 + 5$

$x = 0$

$y = 0 + 0 + 5$

$x = 0$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 5$

$x = 0$

Add $0$ and $5$ .

$y = 5$

$x = 0$

$y = 5$

$x = 0$

$y = 5$

$x = 0$

$y = 5$

$x = 0$

$y = 5$

$x = 0$

Step 4

Replace all occurrences of $x$ with $3$ in each equation.

Replace all occurrences of $x$ in $y = 2 x^{2} - 5 x + 5$ with $3$ .

$y = 2 {(3)}^{2} - 5 \cdot 3 + 5$

$x = 3$

Simplify the right side.

Simplify $2 {(3)}^{2} - 5 \cdot 3 + 5$ .

Simplify each term.

Raise $3$ to the power of $2$ .

$y = 2 \cdot 9 - 5 \cdot 3 + 5$

$x = 3$

Multiply $2$ by $9$ .

$y = 18 - 5 \cdot 3 + 5$

$x = 3$

Multiply $- 5$ by $3$ .

$y = 18 - 15 + 5$

$x = 3$

$y = 18 - 15 + 5$

$x = 3$

Simplify by adding and subtracting.

Subtract $15$ from $18$ .

$y = 3 + 5$

$x = 3$

Add $3$ and $5$ .

$y = 8$

$x = 3$

$y = 8$

$x = 3$

$y = 8$

$x = 3$

$y = 8$

$x = 3$

$y = 8$

$x = 3$

Step 5

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

$(0, 5)$

$(3, 8)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(0, 5), (3, 8)$

Equation Form:

$x = 0, y = 5$

$x = 3, y = 8$

Step 7

Solve the System of Equations y=2x^2-5x+5 y-x=5

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