Solve by Substitution y=x^2-4x+3 2x+y=6

$y = x^{2} - 4 x + 3$ $2 x + y = 6$

Step 1

Replace all occurrences of $y$ with $x^{2} - 4 x + 3$ in each equation.

Replace all occurrences of $y$ in $2 x + y = 6$ with $x^{2} - 4 x + 3$ .

$2 x + x^{2} - 4 x + 3 = 6$

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

Simplify the left side.

Simplify $2 x + x^{2} - 4 x + 3$ .

Remove parentheses.

$2 x + x^{2} - 4 x + 3 = 6$

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

Subtract $4 x$ from $2 x$ .

$x^{2} - 2 x + 3 = 6$

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

$x^{2} - 2 x + 3 = 6$

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

$x^{2} - 2 x + 3 = 6$

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

$x^{2} - 2 x + 3 = 6$

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

Step 2

Solve for $x$ in $x^{2} - 2 x + 3 = 6$ .

Subtract $6$ from both sides of the equation.

$x^{2} - 2 x + 3 - 6 = 0$

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

Subtract $6$ from $3$ .

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

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

Factor $x^{2} - 2 x - 3$ 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 $- 2$ .

$- 3, 1$

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

Write the factored form using these integers.

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

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

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

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

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

$x - 3 = 0$

$x + 1 = 0$

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

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

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

$x - 3 = 0$

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

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 3$

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

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

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

$x + 1 = 0$

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

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = - 1$

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

The final solution is all the values that make $(x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

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

$x = 3, - 1$

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

Step 3

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

Replace all occurrences of $x$ in $y = x^{2} - 4 x + 3$ with $3$ .

$y = {(3)}^{2} - 4 \cdot 3 + 3$

$x = 3$

Simplify the right side.

Simplify ${(3)}^{2} - 4 \cdot 3 + 3$ .

Simplify each term.

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

$y = 9 - 4 \cdot 3 + 3$

$x = 3$

Multiply $- 4$ by $3$ .

$y = 9 - 12 + 3$

$x = 3$

$y = 9 - 12 + 3$

$x = 3$

Simplify by adding and subtracting.

Subtract $12$ from $9$ .

$y = - 3 + 3$

$x = 3$

Add $- 3$ and $3$ .

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

Step 4

Replace all occurrences of $x$ with $- 1$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 4 x + 3$ with $- 1$ .

$y = {(- 1)}^{2} - 4 \cdot - 1 + 3$

$x = - 1$

Simplify the right side.

Simplify ${(- 1)}^{2} - 4 \cdot - 1 + 3$ .

Simplify each term.

Raise $- 1$ to the power of $2$ .

$y = 1 - 4 \cdot - 1 + 3$

$x = - 1$

Multiply $- 4$ by $- 1$ .

$y = 1 + 4 + 3$

$x = - 1$

$y = 1 + 4 + 3$

$x = - 1$

Simplify by adding numbers.

Add $1$ and $4$ .

$y = 5 + 3$

$x = - 1$

Add $5$ and $3$ .

$y = 8$

$x = - 1$

$y = 8$

$x = - 1$

$y = 8$

$x = - 1$

$y = 8$

$x = - 1$

$y = 8$

$x = - 1$

Step 5

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

$(3, 0)$

$(- 1, 8)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, 0), (- 1, 8)$

Equation Form:

$x = 3, y = 0$

$x = - 1, y = 8$

Step 7

Solve by Substitution y=x^2-4x+3 2x+y=6

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