Solve the System of Equations y=x^2-2x+4 y=3x

$y = x^{2} - 2 x + 4$ $y = 3 x$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} - 2 x + 4 = 3 x$

Step 2

Solve $x^{2} - 2 x + 4 = 3 x$ for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $3 x$ from both sides of the equation.

$x^{2} - 2 x + 4 - 3 x = 0$

Subtract $3 x$ from $- 2 x$ .

$x^{2} - 5 x + 4 = 0$

Factor $x^{2} - 5 x + 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $4$ and whose sum is $- 5$ .

$- 4, - 1 + y = 3 x$

Write the factored form using these integers.

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

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

$x - 4 = 0$

$x - 1 = 0 + y = 3 x$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 4, 1$

Step 3

Evaluate $y$ when $x = 4$ .

Substitute $4$ for $x$ .

$y = 3 (4)$

Multiply $3$ by $4$ .

$y = 12$

Step 4

Evaluate $y$ when $x = 1$ .

Substitute $1$ for $x$ .

$y = 3 (1)$

Multiply $3$ by $1$ .

$y = 3$

Step 5

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

$(4, 12)$

$(1, 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 12), (1, 3)$

Equation Form:

$x = 4, y = 12$

$x = 1, y = 3$

Step 7

Solve the System of Equations y=x^2-2x+4 y=3x

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