graph the equation and find the y-intercept.
2y+5x=-6
My question is do I add -5x to both sides first?? how do I start this problem?
When you have an equation with just x and y terms, you're looking at a straight line.
One way to approach is it, as you say, to add -5 to both sides, and then divide across by 2, so that the LHS just contains y on its own. Then you have it in the form y = mx + c, and you can just read off the y-intercept as c.
Another way to find the y-intercept is to set x=0, and see what you're left with:
2y+5x=-6
2y+0 = -6
y=-3
So the point (0,-3) is on the line, and is the y-intercept.
You also want to graph it. Since it's a straight line, you just need two points, then draw the line with a ruler.
Let's get the x-intercept as our other point:
2y+5x=-6
0+5x=-6
x=-6/5
So (-6/5,0) is tyhe x-intercept, and the second point we need for drawing the line.
Oops, typo. In case you're confused, I meant "One way to approach it is, as you say, to add -5x to both sides..."
To graph the equation and find the y-intercept, you need to rearrange the equation into the slope-intercept form (y = mx + b), where m represents the slope, and b represents the y-intercept.
To start, let's rearrange the given equation:
2y + 5x = -6
First, subtract 5x from both sides to isolate the y-term:
2y = -5x - 6
To find the slope-intercept form, divide both sides of the equation by 2:
y = (-5/2)x - 3
Now that we have the equation in slope-intercept form (y = mx + b), we can determine the y-intercept. The y-intercept represents the value of y when x is 0.
In this case, substitute x = 0 into the equation:
y = (-5/2)(0) - 3
y = 0 - 3
y = -3
Therefore, the y-intercept is -3.
Now, to graph the equation, plot the y-intercept (-3) on the y-axis. Then use the slope (-5/2) to find additional points on the graph. The slope tells you how the line "moves" on the graph. For example, if the slope is 3, you would go up 3 units and right 1 unit to find the next point.
Hope that helps! Let me know if you have any more questions.