Question

Can u put the two equations in slope intercept form

Answer 1

To answer this question, we need to rewrite both equations in the slope-intercept form of the line. To achieve this, we can proceed as follows:

1. We need to isolate the variable y in both equations. In the first equation, we need to subtract 3x from both sides of the equation:

`3x+2y=2\Rightarrow3x-3x+2y=2-3x\Rightarrow2y=2-3x`

And now, we need to divide both sides by 2 (to isolate the y variable):

`(2y)/(2)=(2)/(2)-(3x)/(2)\Rightarrow y=1-(3)/(2)x\Rightarrow y=-(3)/(2)x+1`

2. We can proceed in a similar way to find the slope-intercept form of the other line:

`2x-4y=12\Rightarrow2x-2x-4y=12-2x\Rightarrow-4y=12-2x`

Then, we have:

`-(4y)/(4)=(12)/(4)-(2)/(4)x\Rightarrow-y=3-(1)/(2)x\Rightarrow y=(1)/(2)x-3`

Then, we have both line equations in the slope-intercept form.

Now, to graph both equations, we can use the coordinates of the x- and y-intercepts of both lines:

3. The x-intercept is the point when y = 0. Likewise, the y-intercept is the point when x = 0. We need to evaluate the equation in both cases. Then, we have:

First Line

For y = 0:

`0=-(3)/(2)x+1\Rightarrow-1=-(3)/(2)x\Rightarrow1=(3)/(2)x\Rightarrow(2)/(3)\cdot1=(2)/(3)\cdot(3)/(2)x\Rightarrow(2)/(3)=x`

Then, we have the first coordinate: (2/3, 0).

For x = 0:

`y=-(3)/(2)x+1\Rightarrow y=-(3)/(2)(0)+1\Rightarrow y=1`

Then, the other coordinate for this point is (0, 1)

To graph this line we can use these two points (0, 1) and (2/3, 0).

Second Line

We can proceed in a similar way to find these two coordinates for the second line:

For y = 0:

`y=(1)/(2)x-3\Rightarrow0=(1)/(2)x-3\Rightarrow3=(1)/(2)x\Rightarrow2.3=2\cdot(1)/(2)x\Rightarrow6=x\Rightarrow x=6`

The x-intercept is (6, 0).

For x = 0:

`y=(1)/(2)x-3\Rightarrow y=(1)/(2)\cdot(0)-3\Rightarrow y=-3`

The y-intercept is (0, -3).

To graph this line we can use these two points (0, -3) and (6, 0).

Therefore, the solution for this system of equations is x = 2, y = -2.