Find the xx and yy-intercepts of the line 2x+3y=122x+3y=12.

x − i n t : ( 6 , 0 ) x-int:\left(6,0\right) x − in t : ( 6 , 0 ) y − i n t : ( 0 , 4 ) y-int:\left(0,4\right) y − in t : ( 0 , 4 )

Find the x x and y y -intercepts of the line 2 x + 3 y = 12 2x+3y=12 .

This problem asks us to find the x and y intercepts of the line two x plus three y equals 12. So how can we find the x intercept, and how can we find the y intercept? Well, to solve this problem, remember something. When trying to find the x intercept, this is going to happen when you have a y value of zero. And when trying to find the y intercept, this will happen when you have an x value of zero. So let's go over this. First off, what I'm going to do is find the x intercept, and that means setting our y equal to zero. So we're going to have two x plus three times zero is equal to 12. Now three times zero, that whole thing is just going to be equal to zero. So we're going to end up with two x plus zero equals 12. Two x plus zero is the same thing as just two x, and then we divide both sides of this equation by two to isolate that x on one side. 12 over two is equal to six. So x equals six is gonna be the x value for our x intercept. So to find the point that we're looking for, if our x value is six, we know our y value has to be zero, so we get the point six zero, and that would be our x intercept. Now we need to find the y intercept. To find the y intercept, we set the x value equal to zero. Well, doing this, we'll go to this equation here. So we're going to have two times zero plus three y equals 12, and now I'm just gonna be solving for y. Two times zero is zero, so we end up getting zero plus three y equals 12. Zero plus three y is just three y, so that'll equal 12, then I can divide this three on both sides of the equation. Cancelling the threes gives us this y, and 12 over three comes out to just four, so So we get a y value of four. And since we know the x value is zero, our x value is zero, our y value is four, that's our point, and that would be the y intercept. So we have now found the x and y intercepts, and that is the solution to this problem. Hope you found this helpful, and I'll see you in the next one.

Find the x x and y y -intercepts of the line 2 x + 3 y = 12 2x+3y=12 .

x − i n t : ( 6 , 0 ) x-int:\left(6,0\right) x − in t : ( 6 , 0 ) y − i n t : ( 0 , 4 ) y-int:\left(0,4\right) y − in t : ( 0 , 4 )

x − i n t : ( 4 , 0 ) x-int:\left(4,0\right) x − in t : ( 4 , 0 ) y − i n t : ( 0 , 6 ) y-int:\left(0,6\right) y − in t : ( 0 , 6 )

x − i n t : ( 2 , 0 ) x-int:\left(2,0\right) x − in t : ( 2 , 0 ) y − i n t : ( 0 , 3 ) y-int:\left(0,3\right) y − in t : ( 0 , 3 )

x − i n t : ( 0 , 2 ) x-int:\left(0,2\right) x − in t : ( 0 , 2 ) y − i n t : ( 3 , 0 ) y-int:\left(3,0\right) y − in t : ( 3 , 0 )

4. Graphing

Graph Linear Equations Using Intercepts

Beginning Algebra

4. Graphing

Graph Linear Equations Using Intercepts

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Multiple Choice

Find the $x$ and $y$ -intercepts of the line $2 x + 3 y = 12$ .

$x-int:$\left$(4,0$\right$)$

$y-int:$\left$(0,6$\right$)$

$x-int:$\left$(2,0$\right$)$

$y-int:$\left$(0,3$\right$)$

$x-int:$\left$(6,0$\right$)$

$y-int:$\left$(0,4$\right$)$

$x-int:$\left$(0,2$\right$)$

$y-int:$\left$(3,0$\right$)$

Verified step by step guidance

To find the x-intercept of the line given by the equation \$2x + 3y = 12$, start by setting $y = 0$ because the x-intercept occurs where the line crosses the x-axis (where $y$ is zero).

Substitute $y = 0$ into the equation: \$2x + 3(0) = 12$, which simplifies to $2x = 12$.

Solve for $x$ by dividing both sides of the equation by 2: $x = \frac{12}{2}$.

To find the y-intercept, set $x = 0$ because the y-intercept occurs where the line crosses the y-axis (where $x$ is zero).

Substitute $x = 0$ into the original equation: \$2(0) + 3y = 12$, which simplifies to $3y = 12$. Then solve for $y$ by dividing both sides by 3: $y = \frac{12}{3}$.

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