Video: AF5P1-Q17-925160637230

The graph of 𝑦 = 6 βˆ’ 2π‘₯ is shown for the values of π‘₯ from 0 to 3. a) On the same grid, draw the graph of 𝑦 = 3 βˆ’ 1/2 π‘₯ for the values of π‘₯ from 0 to 6. b) Using the graphs, solve the simultaneous equations. 𝑦 = 6 βˆ’ 2π‘₯ and 𝑦 = 3 - 1/2 π‘₯.

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Video Transcript

The graph of 𝑦 equals six minus two π‘₯ is shown for the values of π‘₯ from zero to three. Part a) On the same grid, draw the graph of 𝑦 equals three minus one-half π‘₯ for the values of π‘₯ from zero to six. Part b) Using the graphs, solve the simultaneous equations 𝑦 equals six minus two π‘₯ and 𝑦 equals three minus one-half π‘₯.

Our first task is to draw 𝑦 equals three minus one-half π‘₯ onto the grid. To do this, we’ll need the coordinates of at least two points. However, it’s good to find more than two as that improves the accuracy of our drawing. So we make a table for π‘₯- and 𝑦-values. And we can start by plugging in zero for π‘₯. The equation would say 𝑦 equals three minus one-half times zero. One-half times zero equals zero and three minus zero equals three. When π‘₯ equals zero, 𝑦 equals three. So on our grid, we’ll put a point at π‘₯ equals zero, 𝑦 equals three, the coordinates zero, three.

We need at least one other point. I’ll choose two. I chose to plug in the π‘₯-coordinate two instead of one because we’re multiplying our π‘₯-values by one-half. And that means if we plug in an even number for our π‘₯-value, we’ll get a whole number for our coordinate. So then, we calculate three minus one-half times two. One-half times two is one and three minus one is two. When π‘₯ equals two, 𝑦 equals two. On our grid, we’ll graph the point two, two. Now we could find a third point, where π‘₯ equals four. And so, we solve 𝑦 equals three minus one-half times four. One-half times four equals two and three minus two equals one. When π‘₯ equals four, 𝑦 equals one.

Now we’ll graph our point four, one. If we look closely, we’ll notice that we’re going down one along the 𝑦-axis and right two along the 𝑦-axis. Using this method, we can find our final point down one and right two. We put a point here and we see that when π‘₯ equals six, 𝑦 equals zero. Our question was asking us to draw this graph from zero to six. To do that, we’ll sketch a line across these four points. It’s worth noting here that we could put our functions into 𝑦-intercept form: 𝑦 equals π‘šπ‘₯ plus 𝑐.

The first function we were given that was already graphed is 𝑦 equals six minus two π‘₯. If we rearrange it to 𝑦 equals π‘šπ‘₯ plus 𝑐, it would be 𝑦 equals negative two π‘₯ plus six. If you wanted to graph using slope-intercept form, you start at the 𝑦-intercept. Here that’s six. Since we have a slope of negative two, we can think of this as negative two over one. This tells us we need to go down two and right one.

Our second function 𝑦 equals three minus one-half π‘₯ can be rewritten as 𝑦 equals negative one-half π‘₯ plus three. To graph this function, start at the 𝑦-intercept of three and then we go down one and right two because the slope is negative one-half. And we know that the slope or the gradient is the changes in 𝑦 over the changes in π‘₯. The changes in 𝑦 of this function is negative one and the changes in π‘₯ is positive two β€” down one, right two.

Moving on to part b), using the graph, we want to solve the simultaneous equations. Solving simultaneous equations is finding the coordinate that falls on both lines, the point where they intersect. When we look at our graph, we see that both of these lines have a point at two, two. When π‘₯ equals two, 𝑦 equals two for both of these lines. However, it’s probably good to go ahead and check that this is true.

For our first function 𝑦 equals six minus two π‘₯, we plug in two for our π‘₯- and 𝑦-values. So we have two equals six minus two times two. Two times two is four. Six minus four is two. Two equals two. And that means this statement is true for the first graph. We plug in two for our π‘₯ and 𝑦. And we have two equals three minus one-half times two. One-half times two equals one. Three minus one equals two. And so, the point two, two also lies on our second graph. Using the graph, we found that the simultaneous equations have a solution at two, two.

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