Question Video: Identifying Graphs of Linear Equations in Standard Form | Nagwa Question Video: Identifying Graphs of Linear Equations in Standard Form | Nagwa

Question Video: Identifying Graphs of Linear Equations in Standard Form Mathematics • Second Year of Preparatory School

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Which of the following graphs represents the equation 2π‘₯ βˆ’ 3𝑦 = 18? [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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Video Transcript

Which of the following graphs represents the equation two π‘₯ minus three 𝑦 equals 18?

In this example, we’re given the equation of a linear relation and asked to determine which of the given graphs represents it.

To do this, we recall that if two variables π‘₯ and 𝑦 are related by an equation of the form π‘Žπ‘₯ plus 𝑏𝑦 equals 𝑐, where π‘Ž, 𝑏, and 𝑐 are real constants, then π‘₯ and 𝑦 are linearly related and the relation can be represented by a set of ordered pairs π‘₯, 𝑦. Graphically, this is represented by a straight line. And in our case, the equation is two π‘₯ minus three 𝑦 equals 18. The coordinates π‘₯, 𝑦 of every point on that line are an ordered pair that satisfies the equation. So each point on the line in the correct graph must satisfy the given equation. And if a point on the line does not satisfy the equation, then it cannot be the correct graph.

To determine which graph is correct, we begin by examining what happens in the equation and in each of the graphs when π‘₯ is equal to zero. This will tell us where the line should pass through the 𝑦-axis. Substituting π‘₯ equals zero into the equation and since anything multiplied by zero equals zero, we have negative three 𝑦 is equal to 18. Next, dividing both sides by negative three gives us 𝑦 equal to negative six. This means that the ordered pair zero, negative six satisfies the given equation. And so the point with coordinates π‘₯ equals zero and 𝑦 equals negative six must be on the line in the correct graph. So let’s look at our graphs.

In graph (A), the 𝑦-coordinate of the point on the line that matches with π‘₯ equals zero is 𝑦 equals positive six. Since for a linear equation there is a unique output for each input and the 𝑦-coordinate here is positive six not negative six, the line in graph (A) cannot represent the given equation. So we can eliminate graph (A).

In fact, the same logic applies to graph (D). The 𝑦-coordinate of the point on the line with π‘₯-coordinate equal to zero is again positive six, which doesn’t satisfy the given equation. So at this point, we can also eliminate graph (D).

Now for graph (B), we see that when π‘₯ is equal to zero, the point on the line in this graph has the 𝑦-coordinate negative two. This doesn’t match the 𝑦-coordinate satisfying the equation, which is negative six, so we can also eliminate graph (B).

Looking next at graph (C), when π‘₯ is equal to zero here, the 𝑦-coordinate of the point on the line is 𝑦 equals negative six. And this does match the ordered pair we found, zero, negative six, that satisfies the given equation. So graph (C) is a contender for our linear relation.

We have one graph remaining, graph (E). And we see that when π‘₯ is equal to zero in graph (E), the 𝑦-coordinate of the point on this line is negative nine. This doesn’t satisfy the equation, so we can eliminate graph (E).

So simply by finding a single point on each of the lines in graphs (A), (B), (D), and (E) that does not satisfy the given equation, we can eliminate these graphs. Now to confirm that graph (C), the only remaining graph, really does represent the given equation, we should check that the coordinates of at least one more point on the line satisfy the equation.

Let’s make some space and choose the point on the line in graph (C) with coordinates six, negative two. Substituting these values for π‘₯ and 𝑦 into our equation, we have two times six minus three times negative two. And we want to know if this is equal to 18. Two times six is 12 and negative three times negative two equals six. And adding these together, we do get a total of 18 on the left-hand side. So the point with coordinates six, negative two on the line in graph (C) does satisfy the given equation for the linear relation.

Recalling that there is a unique line through two distinct points in the plane, we can conclude that graph (C) represents the equation two π‘₯ minus three 𝑦 equals 18.

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