Without drawing a graph find the intercepts of each line. y+5=–2x
1 answer:
Answer:
see explanation
Step-by-step explanation:
The intercepts are where the line crosses the x and y axes.
To find the x- intercept ( where the line crosses the x- axis) let y = 0 in the equation and solve for x, that is
0 + 5 = - 2x
5 = - 2x ( divide both sides by - 2 )
x = - = - 2.5 ← x- intercept ⇒ (- 2.5, 0 )
To find the y- intercept ( where the line crosses the y- axis ) let x = 0 in the equation and solve for y, that is
y + 5 = 0 ⇒ y = - 5 ← y- intercept ⇒ (0, - 5 )
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Answer:
C) 2/3x + 3y = 3
Step-by-step explanation:
The attached graph shows only choice C intersects the desired point.
Essentially, the given graph is irrelevant. What you want to know is which equation has (3/4, 5/6) as a solution.
There are at least a couple of ways you can figure this out:
- try the given point in the equations to see which one works
- examine the features of the given equations to see which might work
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If we consider the second approach, we realize all of the equations have negative y-intercepts except choice C*. For an equation with a negative y-intercept to be the solution, the x-intercept must be less than 3/4. That is not the case for any of choices A, B, or D, leaving only choice C as a viable possibility.
We can check to see if the given point satisfies choice C. Filling in for x and y, we get ...
(2/3)(3/4) + 3(5/6) = 2/4 +5/2 = 3 . . . . . . the point is on this line
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In the attached graph, choice C is the black line.
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* For an equation of the form ax+by=c, the y-intercept is c/b and the x-intercept is c/a. The sign of the quotient can be figured from the signs of a, b, and c without doing any division.
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<em>Comment on intercepts</em>
Considering the signs and values of the x- and y-intercepts can tell you a lot about the solution to a system of equations. They tell you where the line segment between the axes lies, and they can give you a clue as to the location (quadrant) of the intersection point of two lines. Often, the intercepts are all you need to create a useful graph of an equation.
