The equations intersecting the point (4, 0) when graphed are x - y = 4, and x + y = 4.
When does a graph intersect a graph?
For any graph y = f(x), to intersect a point (x₁, y₁), the coordinates x₁ and y₁ should satisfy the graph, that is, y₁ need to be equal to f(x₁).
How to solve the question?
In the question, we are asked to determine which of the following equations, when graphed, intersect at the point (4, 0). The equations are:
- x - y = 4
- -x - y = 4
- 2x - y = 7
- x + 4 = 4
- 2x + y = 7
- 2x + y = -7.
We know that for the graphs of the equation, to intersect at the point (4, 0), the coordinates, x-coordinate = 4 and y - coordinate = 0, need to satisfy the equation.
We check for each equation as follows:-
- x - y = 4: We substitute x = 4 and y = 0, to get 4 - 0 = 4, or, 4 = 4, which is true. Hence, the equation x - y = 4, intersects the point (4, 0), when graphed.
- -x - y = 4: We substitute x = 4 and y = 0, to get -4 - 0 = 4, or, -4 = 4, which is false. Hence, the equation -x - y = 4, does not intersect the point (4, 0), when graphed.
- 2x - y = 7: We substitute x = 4 and y = 0, to get 2(4) - 0 = 7, or, 8 - 0 = 7 or, 8 = 7, which is false. Hence, the equation 2x - y = 7, does not intersect the point (4, 0), when graphed.
- x + 4 = 4: We substitute x = 4 and y = 0, to get 4 + 0 = 4, or, 4 = 4, which is true. Hence, the equation x + y = 4, intersects the point (4, 0), when graphed.
- 2x + y = 7: We substitute x = 4 and y = 0, to get 2(4) + 0 = 7, or, 8 + 0 = 7 or, 8 = 7, which is false. Hence, the equation 2x + y = 7, does not intersects the point (4, 0), when graphed.
- 2x + y = -7: We substitute x = 4 and y = 0, to get 2(4) + 0 = -7, or, 8 + 0 = -7 or, 8 = -7, which is false. Hence, the equation 2x + y = -7, does not intersects the point (4, 0), when graphed.
Thus, the equations intersecting the point (4, 0) when graphed are x - y = 4, and x + y = 4.
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