Solve by Substitution x^2-y=0 , 3x+y=10

$x^{2} - y = 0$ , $3 x + y = 10$

Step 1

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

$y = 10 - 3 x$

$x^{2} - y = 0$

Step 2

Replace all occurrences of $y$ with $10 - 3 x$ in each equation.

Replace all occurrences of $y$ in $x^{2} - y = 0$ with $10 - 3 x$ .

$x^{2} - (10 - 3 x) = 0$

$y = 10 - 3 x$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x^{2} - 1 \cdot 10 - (- 3 x) = 0$

$y = 10 - 3 x$

Multiply $- 1$ by $10$ .

$x^{2} - 10 - (- 3 x) = 0$

$y = 10 - 3 x$

Multiply $- 3$ by $- 1$ .

$x^{2} - 10 + 3 x = 0$

$y = 10 - 3 x$

$x^{2} - 10 + 3 x = 0$

$y = 10 - 3 x$

$x^{2} - 10 + 3 x = 0$

$y = 10 - 3 x$

$x^{2} - 10 + 3 x = 0$

$y = 10 - 3 x$

Step 3

Solve for $x$ in $x^{2} - 10 + 3 x = 0$ .

Factor $x^{2} - 10 + 3 x$ 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 $- 10$ and whose sum is $3$ .

$- 2, 5$

$y = 10 - 3 x$

Write the factored form using these integers.

$(x - 2) (x + 5) = 0$

$y = 10 - 3 x$

$(x - 2) (x + 5) = 0$

$y = 10 - 3 x$

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

$x - 2 = 0$

$x + 5 = 0$

$y = 10 - 3 x$

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

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

$x - 2 = 0$

$y = 10 - 3 x$

Add $2$ to both sides of the equation.

$x = 2$

$y = 10 - 3 x$

$x = 2$

$y = 10 - 3 x$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

$y = 10 - 3 x$

Subtract $5$ from both sides of the equation.

$x = - 5$

$y = 10 - 3 x$

$x = - 5$

$y = 10 - 3 x$

The final solution is all the values that make $(x - 2) (x + 5) = 0$ true.

$x = 2, - 5$

$y = 10 - 3 x$

$x = 2, - 5$

$y = 10 - 3 x$

Step 4

Replace all occurrences of $x$ with $2$ in each equation.

Replace all occurrences of $x$ in $y = 10 - 3 x$ with $2$ .

$y = 10 - 3 \cdot 2$

$x = 2$

Simplify the right side.

Simplify $10 - 3 \cdot 2$ .

Multiply $- 3$ by $2$ .

$y = 10 - 6$

$x = 2$

Subtract $6$ from $10$ .

$y = 4$

$x = 2$

$y = 4$

$x = 2$

$y = 4$

$x = 2$

$y = 4$

$x = 2$

Step 5

Replace all occurrences of $x$ with $- 5$ in each equation.

Replace all occurrences of $x$ in $y = 10 - 3 x$ with $- 5$ .

$y = 10 - 3 \cdot - 5$

$x = - 5$

Simplify the right side.

Simplify $10 - 3 \cdot - 5$ .

Multiply $- 3$ by $- 5$ .

$y = 10 + 15$

$x = - 5$

Add $10$ and $15$ .

$y = 25$

$x = - 5$

$y = 25$

$x = - 5$

$y = 25$

$x = - 5$

$y = 25$

$x = - 5$

Step 6

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

$(2, 4)$

$(- 5, 25)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(2, 4), (- 5, 25)$

Equation Form:

$x = 2, y = 4$

$x = - 5, y = 25$

Step 8

Solve by Substitution x^2-y=0 , 3x+y=10

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