QuestionSolve the system:
y = x + 2
5x² + y² = 4
List all solutions as ordered pairs or state if there are none.

Studdy Solution

STEP 1

Assumptions1. We are given a system of two equations - $y = x +$ - $5x^ + y^ =4$ . We are asked to solve this system for $x$ and $y$ .

STEP 2

The first equation is already solved for $y$ . We can substitute $y$ from the first equation into the second equation. This will give us an equation only in terms of $x$ .
$5x^2 + (x +2)^2 =4$

STEP 3

Expand the square in the equation.
$5x^2 + x^2 +x + =$

STEP 4

Combine like terms.
$6x^2 +4x =0$

STEP 5

This is a quadratic equation in the form $ax^2 + bx + c =0$ . We can solve for $x$ by factoring.
$2x(3x +2) =0$

STEP 6

Set each factor equal to zero and solve for $x$ .
$2x =0 \quad \text{and} \quad3x +2 =0$

STEP 7

olving the first equation gives $x =0$ . Solving the second equation gives $x = -\frac{2}{3}$ .

STEP 8

Now that we have the solutions for $x$ , we can substitute these values into the first equation to find the corresponding $y$ values.
For $x =0$ $y =0 +2 =2$ For $x = -\frac{2}{3}$ $y = -\frac{2}{3} +2 = \frac{4}{3}$

STEP 9

So, the solutions to the system of equations are $(,2)$ and $(-\frac{2}{3}, \frac{4}{3})$ .
The solution(s) is/are $(,2)$ and $(-\frac{2}{3}, \frac{4}{3})$ .

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QuestionSolve the system:
y = x + 2
5x² + y² = 4
List all solutions as ordered pairs or state if there are none.

Studdy Solution

STEP 1

Assumptions1. We are given a system of two equations - $y = x +$ - $5x^ + y^ =4$ . We are asked to solve this system for $x$ and $y$ .

STEP 2

The first equation is already solved for $y$ . We can substitute $y$ from the first equation into the second equation. This will give us an equation only in terms of $x$ .
$5x^2 + (x +2)^2 =4$

STEP 3

Expand the square in the equation.
$5x^2 + x^2 +x + =$

STEP 4

Combine like terms.
$6x^2 +4x =0$

STEP 5

This is a quadratic equation in the form $ax^2 + bx + c =0$ . We can solve for $x$ by factoring.
$2x(3x +2) =0$

STEP 6

Set each factor equal to zero and solve for $x$ .
$2x =0 \quad \text{and} \quad3x +2 =0$

STEP 7

olving the first equation gives $x =0$ . Solving the second equation gives $x = -\frac{2}{3}$ .

STEP 8

Now that we have the solutions for $x$ , we can substitute these values into the first equation to find the corresponding $y$ values.
For $x =0$ $y =0 +2 =2$ For $x = -\frac{2}{3}$ $y = -\frac{2}{3} +2 = \frac{4}{3}$

STEP 9

So, the solutions to the system of equations are $(,2)$ and $(-\frac{2}{3}, \frac{4}{3})$ .
The solution(s) is/are $(,2)$ and $(-\frac{2}{3}, \frac{4}{3})$ .

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QuestionSolve the system: y = x + 2 5x² + y² = 4 List all solutions as ordered pairs or state if there are none.

Studdy Solution

STEP 1

Assumptions1. We are given a system of two equations - y=x+y = x +y=x+ - 5x+y=45x^ + y^ =45x+y=4 . We are asked to solve this system for xxx and yyy.

STEP 2

The first equation is already solved for yyy. We can substitute yyy from the first equation into the second equation. This will give us an equation only in terms of xxx.5x2+(x+2)2=45x^2 + (x +2)^2 =45x2+(x+2)2=4

STEP 3

Expand the square in the equation.5x2+x2+x+=5x^2 + x^2 +x + =5x2+x2+x+=

STEP 4

Combine like terms.6x2+4x=06x^2 +4x =06x2+4x=0

STEP 5

This is a quadratic equation in the form ax2+bx+c=0ax^2 + bx + c =0ax2+bx+c=0. We can solve for xxx by factoring.2x(3x+2)=02x(3x +2) =02x(3x+2)=0

STEP 6

Set each factor equal to zero and solve for xxx.2x=0and3x+2=02x =0 \quad \text{and} \quad3x +2 =02x=0and3x+2=0

STEP 7

olving the first equation gives x=0x =0x=0. Solving the second equation gives x=−23x = -\frac{2}{3}x=−32​.

STEP 8

Now that we have the solutions for xxx, we can substitute these values into the first equation to find the corresponding yyy values.For x=0x =0x=0y=0+2=2y =0 +2 =2y=0+2=2For x=−23x = -\frac{2}{3}x=−32​y=−23+2=43y = -\frac{2}{3} +2 = \frac{4}{3}y=−32​+2=34​

STEP 9

So, the solutions to the system of equations are (,2)(,2)(,2) and (−23,43)(-\frac{2}{3}, \frac{4}{3})(−32​,34​).The solution(s) is/are (,2)(,2)(,2) and (−23,43)(-\frac{2}{3}, \frac{4}{3})(−32​,34​).

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QuestionSolve the system:
y = x + 2
5x² + y² = 4
List all solutions as ordered pairs or state if there are none.

Assumptions1. We are given a system of two equations - $y = x +$ - $5x^ + y^ =4$ . We are asked to solve this system for $x$ and $y$ .

The first equation is already solved for $y$ . We can substitute $y$ from the first equation into the second equation. This will give us an equation only in terms of $x$ .
$5x^2 + (x +2)^2 =4$

Expand the square in the equation.
$5x^2 + x^2 +x + =$

Combine like terms.
$6x^2 +4x =0$

This is a quadratic equation in the form $ax^2 + bx + c =0$ . We can solve for $x$ by factoring.
$2x(3x +2) =0$

Set each factor equal to zero and solve for $x$ .
$2x =0 \quad \text{and} \quad3x +2 =0$

olving the first equation gives $x =0$ . Solving the second equation gives $x = -\frac{2}{3}$ .

Now that we have the solutions for $x$ , we can substitute these values into the first equation to find the corresponding $y$ values.
For $x =0$ $y =0 +2 =2$ For $x = -\frac{2}{3}$ $y = -\frac{2}{3} +2 = \frac{4}{3}$

So, the solutions to the system of equations are $(,2)$ and $(-\frac{2}{3}, \frac{4}{3})$ .
The solution(s) is/are $(,2)$ and $(-\frac{2}{3}, \frac{4}{3})$ .