Graphing Linear Equations: Free Response

Substitute each pair into $4x - y = 6$, putting the first coordinate in for $x$ and the second for $y$, then compare the two sides.

For $(3, 6)$:

$4(3) - 6 = 12 - 6 = 6,$

which is exactly the $6$ the equation demands, so $(3, 6)$ is on the line.

For $(1, -2)$, the subtraction of a negative $y$-value is where a sign usually slips:

$4(1) - (-2) = 4 + 2 = 6,$

so $(1, -2)$ is on the line as well.

For $(0, 6)$:

$4(0) - 6 = 0 - 6 = -6,$

and $-6$ is not $6$, so $(0, 6)$ is off the line. Compare it with the first point: $(3, 6)$ and $(0, 6)$ share a $y$-coordinate, and sharing one coordinate with a point of the line settles nothing at all.

A point lies on the line exactly when its coordinates satisfy the equation. Each of these points hands you one coordinate, and the equation then supplies the other.

For $(k, 10)$, put $y = 10$ into $4x - y = 6$ and solve for the $x$-coordinate:

$4k - 10 = 6, \qquad 4k = 16, \qquad k = 4,$

so that point is $(4, 10)$.

For the point whose $y$-coordinate is $1$, put $y = 1$ in and solve the same way:

$4x - 1 = 6, \qquad 4x = 7, \qquad x = \tfrac{7}{4},$

so that point is $\left(\tfrac{7}{4}, 1\right)$. The fraction is not a symptom of a mistake. Between the corners of the grid the line is packed with solutions whose coordinates are not whole numbers, and this is one of them.

Rearranging is not a change of subject. It is a change of clothes.

Subtract $4x$ from both sides of $4x - y = 6$, then multiply through by $-1$:

$-y = 6 - 4x, \qquad y = 4x - 6.$

Every step there is reversible, so a pair $(x, y)$ satisfies one equation exactly when it satisfies the other: the two equations have precisely the same solutions. The graph of an equation is nothing but its set of solutions, so two equations with the same solutions have the same graph, down to the last point. The picture cannot notice which form you happened to write.

What the rearrangement does change is the labour. In $4x - y = 6$ a chosen $x$ still leaves an equation to be solved for $y$; in $y = 4x - 6$ the value of $y$ is computed outright:

$x = 3: \; y = 4(3) - 6 = 6.$

That is exactly what a table of values wants, one row at a time, and it is also what part A's three checks become: compute $4x - 6$ and see whether it agrees with the $y$ you were handed.

Rearranging first is what makes a table quick: do the algebra once, then only arithmetic.

Subtract $x$ from both sides of $x + 2y = 6$ and divide by $2$:

$2y = 6 - x, \qquad y = \tfrac{6 - x}{2} = 3 - \tfrac{x}{2}.$

Now compute one $y$ for each chosen $x$:

$x = -2: \; y = 3 + 1 = 4; \qquad x = 0: \; y = 3 - 0 = 3.$

$x = 2: \; y = 3 - 1 = 2; \qquad x = 4: \; y = 3 - 2 = 1.$

So the four solutions are $(-2, 4)$, $(0, 3)$, $(2, 2)$ and $(4, 1)$.

The choice of $x$-values was not innocent. The rearranged form halves $x$, so an even $x$ leaves a whole number behind while an odd one would leave a half. Taking $-2$, $0$, $2$ and $4$ therefore lands every point on a corner of the grid, and it also spreads the points either side of zero instead of crowding them into one corner.

Each intercept is found by zeroing the coordinate of the axis being crossed, which is always the opposite letter to the one you are looking for.

Every point of the x-axis has $y = 0$, so set $y = 0$ in $4x - 3y = 12$:

$4x - 3(0) = 12, \qquad 4x = 12, \qquad x = 3,$

and the x-intercept is $(3, 0)$.

Every point of the y-axis has $x = 0$, so set $x = 0$:

$4(0) - 3y = 12, \qquad -3y = 12, \qquad y = -4,$

and the y-intercept is $(0, -4)$. A negative intercept is nothing unusual; it simply puts that crossing below the origin. Plot $(3, 0)$ and $(0, -4)$, draw the line through them, and the graph is finished.

Both routes do the same thing, and it is worth saying what.

A table row is a solution: choose $x$, then compute the $y$ that makes the equation true. An intercept is a solution too: choose the value $0$ for one of the two letters, then compute the other. The intercept route does not produce a different kind of point, only a different way of choosing which solutions to compute, picked because zero makes one term vanish and because the resulting points sit on the axes, where they are easy to plot accurately.

That also settles where a fifth solution of $x + 2y = 6$ has to land. The graph of an equation is its whole set of solutions, and for a linear equation that set is a straight line, so any further solution is a point of the same line as the four already plotted. Take $x = 6$:

$y = 3 - \tfrac{6}{2} = 0, \qquad \text{so } (6, 0),$

which continues the same line rather than starting a new one.

And if one of the four points had sat apart from the rest, the graph would not be to blame. A linear equation's graph does not bend, so the fault would lie in that row's arithmetic. Recompute the row and the point rejoins the others, which is exactly why plotting a third and fourth solution is worth the trouble.

Say the situation in words first, then in symbols.

After $t$ minutes the pump has removed $6t$ litres. Those litres plus the litres still in the tank make up the $48$ the tank started with, so

$6t + v = 48.$

Solving for $v$ says the same thing in the form a table would want:

$v = 48 - 6t.$

What makes this an equation of the graphable kind is how the letters appear in it. Each of $t$ and $v$ stands alone, to the first power, multiplied only by a number, with no product of the two letters and no power above the first. That is exactly the shape the lesson calls linear, and it is why the solution pairs will fall on one straight line rather than scattering or curving.

The two intercepts come from the same move as always: zero the coordinate belonging to the axis you are crossing.

Every point of the vertical axis has $t = 0$, so set $t = 0$ in $6t + v = 48$:

$6(0) + v = 48, \qquad v = 48,$

which gives the intercept $(0, 48)$.

Every point of the horizontal axis has $v = 0$, so set $v = 0$:

$6t + 0 = 48, \qquad 6t = 48, \qquad t = 8,$

which gives the intercept $(8, 0)$. Plotting those two points and drawing the line through them graphs the whole equation.

Read each pair back as a sentence, with its units.

The intercept $(0, 48)$ has $t = 0$, the moment the pump was switched on, and $v = 48$: at that instant the tank held $48$ litres, which is the tank full. The intercept $(8, 0)$ has $v = 0$, no water left, at $t = 8$: the tank ran dry $8$ minutes after the pump started. So the two intercepts are not merely convenient points here. They are the beginning and the end of the story.

The pair $(5, 18)$ was computed by nobody, and it is on the line all the same:

$6(5) + 18 = 30 + 18 = 48.$

That is what the completed line adds. The two intercepts answer two questions; the line answers all of them at once, because it is the entire solution set drawn in one stroke. To find how much water is left after five minutes you can go up from $t = 5$ and read across, and to find when $18$ litres remain you can go across from $v = 18$ and read down, without solving anything on paper.

A claim about every line is sunk by one line, so the job is to produce one and check it, not to argue in general.

Take $y = 3$. The lesson graphs it as a horizontal line three units above the x-axis, and it is a linear equation like any other: written out in full it is $0 \cdot x + y = 3$, with the coefficient of $x$ equal to zero.

Now do what the claim promises and set $y = 0$ to find an x-intercept:

$0 = 3.$

That statement is false, and no choice of $x$ can change it, because $x$ has already vanished from the equation. So no solution of $y = 3$ has $y$-coordinate $0$, no point of its graph lies on the x-axis, and the line has no x-intercept whatsoever. One line is enough, so the claim is false as stated.

The vertical family fails in the mirror-image way. Setting $x = 0$ in $x = -2$ gives

$0 = -2,$

so that line has no y-intercept.

Repairing a claim means naming a family and then arguing both halves: that every line inside it qualifies, and that every line outside it does not.

The family is the lines that are neither horizontal nor vertical. Write such a line as $ax + by = c$ with neither $a$ nor $b$ equal to zero, which is exactly what being neither horizontal nor vertical amounts to: a zero coefficient on $x$ leaves $y$ alone and pins the height, and a zero coefficient on $y$ pins the column.

Setting $y = 0$ leaves $ax = c$, and since $a$ is not zero that has exactly one answer:

$x = \tfrac{c}{a}.$

Setting $x = 0$ leaves $by = c$, and since $b$ is not zero that has exactly one answer too:

$y = \tfrac{c}{b}.$

So every line of the family has exactly one x-intercept and exactly one y-intercept.

Now the lines left out, and each family needs two cases rather than one. A horizontal line is $y = k$. If $k$ is not zero, setting $y = 0$ demands $0 = k$, which is false, so there is no x-intercept. If $k$ is zero, the line IS the x-axis, and then every one of its infinitely many points lies on the x-axis, which is not exactly one either. A vertical line $x = k$ fails the same two ways against the y-axis.

Every line is horizontal, vertical, or neither, so the cases are exhaustive and both directions are covered. The repaired claim is therefore exact: a line has exactly one x-intercept and exactly one y-intercept precisely when it is neither horizontal nor vertical.

The repaired claim of part B is about existence, but drawing a line asks for more than existence: it asks for two DIFFERENT points.

Take a line of the family that passes through the origin, such as $y = 2x$. Setting $y = 0$ gives

$2x = 0, \qquad x = 0,$

so the x-intercept is $(0, 0)$. Setting $x = 0$ gives

$y = 2(0) = 0,$

so the y-intercept is $(0, 0)$ as well.

The two computations land on the same point, and they had to. An x-intercept lies on the x-axis and a y-intercept lies on the y-axis, so the only way for them to be the same point is for that point to lie on both axes, and the origin is the only point that does. That is why the collapse happens for exactly the lines through the origin and for no others.

One point does not fix a line, since infinitely many lines pass through any single point. The repair is to give up on the second intercept and compute an ordinary solution instead: choose any value of $x$ other than $0$ and work out its $y$. For $y = 2x$, choosing $x = 3$ gives

$y = 2(3) = 6,$

so $(3, 6)$ joins the origin, and the line through those two points is the graph.

The picture is not the authority here. The equation is, so substitute each point into it.

For $P = (4, 3)$:

$3(4) + 4(3) = 12 + 12 = 24,$

which is exactly what the equation demands, so $P$ is a solution and lies on the graph.

For $Q = (5, 2)$:

$3(5) + 4(2) = 15 + 8 = 23,$

and $23$ is not $24$, so $Q$ is not a solution and is not on the graph, however close the drawing puts it. The two sides differ by $1$, and on this line that corresponds to $Q$ sitting a quarter of a unit below it, which is smaller than the width of a drawn line and a plotted dot together.

Two facts are doing the work here, and neither of them is about these particular points.

First, the graph of a linear equation is a straight line. Second, through two distinct points there passes exactly one straight line.

Now run the argument. Both intercepts are solutions, so both are points of the graph. For $3x + 4y = 24$ they are $(8, 0)$ and $(0, 6)$, and substituting confirms it:

$3(8) + 4(0) = 24, \qquad 3(0) + 4(6) = 24.$

The graph is a straight line containing those two distinct points. Only one straight line does that, and it is the one drawn through them. So the drawn line IS the graph, and since a point of the graph is by definition a solution, every one of the line's infinitely many points is a solution, although only two of them were ever tested.

Notice how much each fact is carrying. Drop the first and the graph might be some curve running through the two points, so the line drawn would be a different set entirely. Drop the second and the two points would pin nothing down, since many lines could pass through them. Two points are enough only because both facts hold at once.

The two are not rivals, because they answer different questions.

Substitution is decisive. The graph is defined as the set of pairs that satisfy the equation, so testing a pair against the equation tests exactly the thing at issue, and it returns a verdict with no margin at all:

$3(5) + 4(2) = 23, \qquad 23 \neq 24.$

Its limit is its scope. It settles the one point you feed it, and a line has infinitely many.

The drawing is the opposite on both counts. It shows the entire solution set at once, which no substitution can, and it is what raises the question in the first place: nobody would have thought to test $Q$ at all if the picture had not put it against the line. But it settles nothing exactly. A drawn line has a width, a plotted dot has a radius, and part A's gap of a quarter of a unit is smaller than the width of a drawn line and a plotted dot together.

So part B does not retire the drawing. It licenses it. Part B is precisely the guarantee that a line drawn through two solutions really does display every solution and nothing else, which is what makes the picture worth reading at all. The division of labour is that the picture proposes and the equation disposes.