Solve the System of Equations y=x^2+2 y=4x-1

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

Step 1

Eliminate the equal sides of each equation and combine.

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

Step 2

Solve $x^{2} + 2 = 4 x - 1$ for $x$ .

Subtract $4 x$ from both sides of the equation.

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

Add $1$ to both sides of the equation.

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

Add $2$ and $1$ .

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

Factor $x^{2} - 4 x + 3$ 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 $3$ and whose sum is $- 4$ .

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

Write the factored form using these integers.

$(x - 3) (x - 1) = 0$

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

$x - 3 = 0$

$x - 1 = 0 + y = 4 x - 1$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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 $(x - 3) (x - 1) = 0$ true.

$x = 3, 1$

Step 3

Evaluate $y$ when $x = 3$ .

Substitute $3$ for $x$ .

$y = 4 (3) - 1$

Simplify $4 (3) - 1$ .

Multiply $4$ by $3$ .

$y = 12 - 1$

Subtract $1$ from $12$ .

$y = 11$

Step 4

Evaluate $y$ when $x = 1$ .

Substitute $1$ for $x$ .

$y = 4 (1) - 1$

Simplify $4 (1) - 1$ .

Multiply $4$ by $1$ .

$y = 4 - 1$

Subtract $1$ from $4$ .

$y = 3$

Step 5

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

$(3, 11)$

$(1, 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, 11), (1, 3)$

Equation Form:

$x = 3, y = 11$

$x = 1, y = 3$

Step 7

Solve the System of Equations y=x^2+2 y=4x-1

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