Algebra Examples

y = 1

$y = 1$ ,

y = x + 3

$y = x + 3$

Step 1

Use the slope-intercept form to find the slope.

Tap for more steps...

Step 1.1

The slope-intercept form is

y = m x + b

$y = m x + b$ , where

m

$m$ is the slope and

b

$b$ is the y-intercept.

y = m x + b

$y = m x + b$

Step 1.2

Using the slope-intercept form, the slope is

0

$0$ .

m_{1} = 0

$m_{1} = 0$

m_{1} = 0

$m_{1} = 0$

Step 2

Use the slope-intercept form to find the slope.

Tap for more steps...

Step 2.1

The slope-intercept form is

y = m x + b

$y = m x + b$ , where

m

$m$ is the slope and

b

$b$ is the y-intercept.

y = m x + b

$y = m x + b$

Step 2.2

Using the slope-intercept form, the slope is

1

$1$ .

m_{2} = 1

$m_{2} = 1$

m_{2} = 1

$m_{2} = 1$

Step 3

Set up the system of equations to find any points of intersection.

y = 1, y = x + 3

$y = 1, y = x + 3$

Step 4

Solve the system of equations to find the point of intersection.

Tap for more steps...

Step 4.1

Eliminate the equal sides of each equation and combine.

1 = x + 3

$1 = x + 3$

Step 4.2

Solve

1 = x + 3

$1 = x + 3$ for

x

$x$ .

Tap for more steps...

Step 4.2.1

Rewrite the equation as

x + 3 = 1

$x + 3 = 1$ .

x + 3 = 1

$x + 3 = 1$

Step 4.2.2

Move all terms not containing

x

$x$ to the right side of the equation.

Tap for more steps...

Step 4.2.2.1

Subtract

3

$3$ from both sides of the equation.

x = 1 - 3

$x = 1 - 3$

Step 4.2.2.2

Subtract

3

$3$ from

1

$1$ .

x = - 2

$x = - 2$

x = - 2

$x = - 2$

x = - 2

$x = - 2$

Step 4.3

Evaluate

y

$y$ when

x = - 2

$x = - 2$ .

Tap for more steps...

Step 4.3.1

Substitute

- 2

$- 2$ for

x

$x$ .

y = (- 2) + 3

$y = (- 2) + 3$

Step 4.3.2

Substitute

- 2

$- 2$ for

x

$x$ in

y = (- 2) + 3

$y = (- 2) + 3$ and solve for

y

$y$ .

Tap for more steps...

Step 4.3.2.1

Remove parentheses.

y = - 2 + 3

$y = - 2 + 3$

Step 4.3.2.2

Remove parentheses.

y = (- 2) + 3

$y = (- 2) + 3$

Step 4.3.2.3

Add

- 2

$- 2$ and

3

$3$ .

y = 1

$y = 1$

y = 1

$y = 1$

y = 1

$y = 1$

Step 4.4

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

(- 2, 1)

$(- 2, 1)$

(- 2, 1)

$(- 2, 1)$

Step 5

Since the slopes are different, the lines will have exactly one intersection point.

m_{1} = 0

$m_{1} = 0$

m_{2} = 1

$m_{2} = 1$

(- 2, 1)

$(- 2, 1)$

Step 6

Enter YOUR Problem