QuanticaAlgebra IOpen in the course7.1 gave every ordered pair a place on the plane. From 5.1, an equation in two variables has one solution pair for every value of \(x\) chosen, so its solutions can never be listed in full. Every one of those pairs is a point. Marking all of them at once replaces the endless list with one picture.
Problem
The graph of \(3x+4y=29\) is the set of points whose coordinates make that equation true. One of \((7,2)\) and \((9,1)\) is on the graph and the other is not. Substitute both, and enter the value the left side takes at the point that is not on the graph.
Show a hint
- Substituting a point puts its first coordinate in for \(x\) and its second for \(y\), then evaluates \(3x+4y\). Do that for both points and compare each result with \(29\).
- Start with \((7,2)\). If \(3(7)+4(2)\) comes out to \(29\), that point is on the graph, and the value you want is whatever \(3x+4y\) gives at \((9,1)\).
Show the full solution
Substituting \((9,1)\) into the left side gives \(3(9)+4(1)=27+4=\boxed{31}\). The other point does satisfy the equation, since \(3(7)+4(2)=21+8=29\), so \((7,2)\) is on the graph and \((9,1)\), where the left side comes out to \(31\) instead of \(29\), is not.
Problem
The notation \((x,8)\) means a point whose second coordinate is \(8\) and whose first coordinate is not yet known. That point is on the graph of \(5x-3y=11\). Enter its \(x\)-coordinate.
Show a hint
- The number given is the second coordinate, so it is the value of \(y\). Substitute it and the equation has only \(x\) left in it.
- The equation becomes \(5x-24=11\).
Show the full solution
Substituting \(y=8\) gives \(5x-3(8)=11\), which is \(5x-24=11\). Then \(5x=35\) and \(x=\boxed{7}\). A point is on the graph exactly when its coordinates satisfy the equation, so the known coordinate goes in for its own letter and the missing one is left to solve for.
Problem
Build a table for \(3x+2y=30\) at \(x=0,\ 2,\ 4,\ 6,\ 8\), solving for \(y\) in each row. The five pairs all lie on one straight line, and exactly one of them has its two coordinates equal. Enter the value those two coordinates share.
Show a hint
- Each row takes one value of \(x\) and solves \(3x+2y=30\) for \(y\), so the row for \(x=2\) reads \(6+2y=30\).
- A row with equal coordinates has \(y\) equal to its own \(x\), so replacing \(y\) with \(x\) in \(3x+2y=30\) turns it into an equation in \(x\) alone.
Show the full solution
The rows give \((0,15)\), \((2,12)\), \((4,9)\), \((6,6)\) and \((8,3)\), and the pair with equal coordinates is \((6,6)\), so the shared value is \(\boxed{6}\). All five pairs satisfy the equation, so all five are points of the same line, and any two of them are enough to draw it.
Problem
Exactly one of these four points is not on the graph of \(4x-3y=18\). $$\left(\tfrac{3}{4},-5\right)\qquad(3,-2)\qquad(9,7)\qquad(12,10)$$ Enter the \(x\)-coordinate of the point that is not.
Show a hint
- Test all four the same way, including the one with a fractional coordinate. A fraction is a perfectly good coordinate, and all that matters is whether \(4x-3y\) comes out to \(18\).
- Watch the signs. At \(\left(\tfrac34,-5\right)\) the term \(-3y\) becomes \(+15\), and at \((3,-2)\) it becomes \(+6\).
Show the full solution
The four left sides are \(3+15=18\), \(12+6=18\), \(36-21=15\) and \(48-30=18\). Only \((9,7)\) gives something other than \(18\), and its \(x\)-coordinate is \(\boxed{9}\). Whether a point is on the graph depends only on what the substitution gives, not on how tidy the coordinates look, so here the fractional point is on the line and a whole-number point is not.
Two points are enough to draw the line, so a third pair is spare, and a spare point is a check. Any two points lie in a straight line, even two computed wrongly, so an arithmetic slip is invisible with only two points. Three points in one line is a condition that can fail, so a third point off the line is proof that one of the three is wrong.
Problem
The point \((4,2)\) is on the graph of \(7x-ky=6\), where \(k\) is a constant. Only one value of \(k\) makes that true. Enter \(k\).
Show a hint
- Testing a point is the same substitution whether the unknown is a coordinate or a coefficient. Putting both coordinates in leaves an equation whose only unknown is \(k\).
- With \(x=4\) and \(y=2\) the equation reads \(28-2k=6\). Note the minus sign in front of the \(k\) term.
Show the full solution
Substituting gives \(7(4)-k(2)=6\), so \(28-2k=6\), then \(2k=22\) and \(k=\boxed{11}\). Check, \(7(4)-11(2)=28-22=6\). The usual slip is reading \(-k(2)\) as \(+2k\), which gives \(28+2k=6\) and \(k=-11\).
Problem
A student computes three pairs for \(2x+5y=34\) and gets \((2,6)\), \((7,4)\) and \((12,3)\). Plotted, the three points do not fall in a straight line, so one pair is wrong. Keep the \(x\)-value of the pair that fails, and enter the \(y\)-value that belongs with it.
Show a hint
- Two of the three pairs satisfy the equation and one does not, so run the test on all three before deciding which one to fix.
- The three left sides come out \(34\), \(34\) and \(39\). Take the \(x\)-value of the pair that misses and solve \(2x+5y=34\) for \(y\) again.
Show the full solution
The three left sides are \(4+30=34\), \(14+20=34\) and \(24+15=39\), so \((12,3)\) is the wrong pair. At \(x=12\), \(5y=34-24=10\) and \(y=\boxed{2}\). The corrected point sits one unit below the plotted one, which is why the three looked nearly straight rather than obviously wrong. A third point is worth computing for exactly that reason.
Any value of \(x\) is allowed in the table, and two choices take less work than the rest. Where a graph crosses an axis, one coordinate is already known, so only the other is left to compute. The next two problems find those crossings, and a later pair shows that a graph need not cross both axes at all.
Problem
The graph of \(3x+8y=24\) is a straight line, and it meets the \(x\)-axis at exactly one point. Enter the \(x\)-coordinate of that point.
Show a hint
- Every point of the \(x\)-axis has \(y=0\), which 7.1 established. That is the value to substitute here.
- With \(y=0\) the term \(8y\) is \(0\), so the equation becomes \(3x=24\).
Show the full solution
A point of the \(x\)-axis has \(y=0\), so \(3x+8(0)=24\), which is \(3x=24\), and \(x=\boxed{8}\). Setting \(x=0\) instead gives \(8y=24\) and \(y=3\), the crossing on the \(y\)-axis rather than the \(x\)-axis.
Problem
The graph of \(3x-4y=48\) is a straight line that crosses the \(y\)-axis once. Enter the \(y\)-coordinate of the crossing point.
Show a hint
- Every point on the \(y\)-axis has \(x=0\), so substitute that value and solve for \(y\).
- The equation becomes \(-4y=48\), and the coefficient of \(y\) is negative, so divide both sides by \(-4\) and keep the sign.
Show the full solution
A point on the \(y\)-axis has \(x=0\), so \(3(0)-4y=48\), which is \(-4y=48\), and \(y=\boxed{-12}\). Dropping the minus gives \(12\) and puts the crossing above the origin instead of below it, and setting \(y=0\) by mistake gives \(16\), which is the \(x\)-axis crossing.
Every equation so far has had both letters in it. In \(ax+by=c\) only one of \(a\) and \(b\) has to be nonzero, so the other may be zero. With \(a=0\) what remains is \(by=c\), and dividing by \(b\) leaves \(y=k\). With \(b=0\) what remains is \(x=k\). Each still has a graph, and the next two problems settle which points are on it.
Problem
In the coordinate plane, the graph of \(2y+9=-5\) is made of exactly the points \((x,y)\) that satisfy it. One of those points has first coordinate \(40\). Enter that point's second coordinate.
Show a hint
- The equation contains no \(x\), so the first coordinate can be anything at all. The only condition is on the second coordinate.
- Solve \(2y+9=-5\) for \(y\) the same way you would solve any one-variable equation. The \(40\) is not needed for that work.
Show the full solution
From \(2y+9=-5\), \(2y=-14\) and \(y=\boxed{-7}\). Because the equation contains no \(x\), every point on this graph has second coordinate \(-7\), whatever the first coordinate is. That is why the \(40\) is not needed.
Problem
The graph of \(3x=-18\) is a straight line. Enter the number of points that line has in common with the \(y\)-axis.
Show a hint
- Solve for \(x\) first, then describe the graph. No \(y\) appears, so every point of the graph shares one \(x\)-coordinate.
- From 7.1, a point is on the \(y\)-axis exactly when its \(x\)-coordinate is \(0\). Compare that with the \(x\)-coordinate every point of this graph has.
Show the full solution
Solving gives \(x=-6\), so every point of the graph has first coordinate \(-6\). A point of the \(y\)-axis has first coordinate \(0\), and \(-6\neq 0\), so the number of shared points is \(\boxed{0}\). Reading \(x=-6\) as a horizontal line gives the answer \(1\) instead, since a horizontal line does meet the \(y\)-axis once.
The last two problems use every piece at once. An equation is not always written in the form \(ax+by=c\), and terms in the same letter may appear on both sides, so collecting them onto one side comes first. A coefficient can also be unknown, and then one stated fact about the graph is enough to determine it.
Problem
The graph of \(9y-4x=30+5y\) is a straight line. Enter the \(x\)-coordinate of the point where it meets the \(x\)-axis.
Show a hint
- The \(y\) terms sit on both sides. Collect them on one side first, and the equation takes the form \(ax+by=c\).
- Collecting gives \(4y-4x=30\). Every point of the \(x\)-axis has \(y=0\), so put that value in and solve for \(x\).
Show the full solution
Subtracting \(5y\) from both sides gives \(4y-4x=30\). A point of the \(x\)-axis has \(y=0\), so \(-4x=30\) and \(x=\boxed{-\frac{15}{2}}\). Setting \(x=0\) instead gives \(4y=30\) and \(y=\frac{15}{2}\), which is where the line meets the \(y\)-axis, not the \(x\)-axis.
Problem
The graph of \(6x+by=21\) has no point in common with the \(y\)-axis, where \(b\) is a constant. Enter the \(x\)-coordinate of the point on that graph whose \(y\)-coordinate is \(-11\).
Show a hint
- Substituting \(y=-11\) straight away leaves two unknowns, so use the stated fact first. A point sits on the \(y\)-axis exactly when its \(x\)-coordinate is \(0\), so put \(x=0\) into the equation.
- If \(b\neq0\), then \(x=0\) gives \(by=21\) and a point on the \(y\)-axis exists. The graph has no such point, so only one value of \(b\) is possible, and the equation is left with no \(y\) term.
Show the full solution
If \(b\neq 0\), setting \(x=0\) gives \(by=21\) and a point on the \(y\)-axis. The graph has no such point, so \(b=0\) and the equation is \(6x=21\). Every point of that graph has \(x=\frac{21}{6}\), so the one with \(y=-11\) has \(x=\boxed{\frac{7}{2}}\). The form \(ax+by=c\) allows \(b=0\) as long as \(a\) is not zero too, and here \(a=6\). With \(b=0\) the equation contains no \(y\), so the \(-11\) is not needed.
Every graph in this lesson came from the same two steps, find two solution points and draw the line through them. Those steps settle where the line is and say nothing about how steep it is. Lesson 7.3, Slope, attaches a number to the steepness, computed from any two points of the line.
Practice these ideas
Practice
Substitute the point \((6,3)\) into the left side of \(2x+3y=19\), and enter the resulting value of \(2x+3y\).
Show the solution
Substituting gives \(2(6)+3(3)=12+9=\boxed{21}\). The left side comes out \(21\) while the right side is \(19\), so \((6,3)\) does not satisfy the equation and does not lie on the graph.
Practice
The point \((4,y)\) is on the graph of \(x+y=17\). Enter \(y\).
Show the solution
The point has \(x=4\), so \(4+y=17\) and \(y=\boxed{13}\). A point is on the graph exactly when its two coordinates make the equation true, so putting in the coordinate that is given leaves one equation in one unknown.
Practice
The point \((x,3)\) is on the graph of \(5x-2y=64\). Enter the value of \(x\).
Show the solution
The point has \(y=3\), so \(5x-2(3)=64\), which is \(5x-6=64\). Then \(5x=70\) and \(x=\boxed{14}\). Check, \(5(14)-2(3)=70-6=64\). Putting the \(3\) in for \(x\) instead gives \(15-2y=64\), a different point on the same line.
Practice
The graph of \(9x+2y=-8\) contains a point whose \(x\)-coordinate is \(-2\). Enter that point's \(y\)-coordinate.
Show the solution
Substituting \(x=-2\) gives \(9(-2)+2y=-8\), so \(-18+2y=-8\), then \(2y=10\) and \(y=\boxed{5}\). Dropping the negative gives \(18+2y=-8\) and \(y=-13\), the common wrong answer.
Practice
Three of \((5,13)\), \((8,7)\), \((10,4)\), \((1,21)\) are on the graph of \(2x+y=23\) and one is not. Enter the \(x\)-coordinate of the one that is not.
Show the solution
Substituting gives \(10+13=23\), \(16+7=23\), \(20+4=24\) and \(2+21=23\). The third value is not \(23\), so the point off the graph is \((10,4)\) and the answer is \(\boxed{10}\). A point is on the graph exactly when its coordinates satisfy the equation, so one failed check is enough to rule a point out.
Practice
Enter the \(x\)-coordinate of the point where the graph of \(5x+2y=60\) meets the \(x\)-axis.
Show the solution
Every point on the \(x\)-axis has \(y=0\), so \(5x+2(0)=60\), which is \(5x=60\), and \(x=\boxed{12}\). Reversing the substitution and setting \(x=0\) gives \(2y=60\) and \(y=30\), which is the \(y\)-coordinate of the point where the graph meets the \(y\)-axis.
Practice
Enter the \(y\)-coordinate of the point where the graph of \(7x-3y=42\) meets the \(y\)-axis.
Show the solution
Every point on the \(y\)-axis has \(x=0\), so \(7(0)-3y=42\), which is \(-3y=42\), and \(y=\boxed{-14}\). Setting \(y=0\) instead gives \(7x=42\) and the number \(6\), which is where the graph meets the \(x\)-axis, not the \(y\)-axis.
Practice
A point on the graph of \(5y+31=6\) has first coordinate \(-90\). Enter its second coordinate.
Show the solution
From \(5y+31=6\), \(5y=-25\) and \(y=\boxed{-5}\). The equation contains no \(x\), so every point of this graph has second coordinate \(-5\) whatever its first coordinate is, which makes the \(-90\) irrelevant.
Practice
In the coordinate plane, the graph of \(4x+29=5\) contains a point whose second coordinate is \(41\). Enter that point's first coordinate.
Show the solution
From \(4x+29=5\), subtracting \(29\) gives \(4x=-24\), so \(x=\boxed{-6}\). The equation has no \(y\) in it, so every point of the graph has first coordinate \(-6\) whatever its second coordinate is, and the graph is the vertical line \(x=-6\). The \(41\) is not needed.
Practice
The point \((2,5)\) is on the graph of \(kx-4y=10\), where \(k\) is a constant. Enter \(k\).
Show the solution
Substituting gives \(k(2)-4(5)=10\), so \(2k-20=10\), then \(2k=30\) and \(k=\boxed{15}\). Checking it back, \(15(2)-4(5)=30-20=10\), so \((2,5)\) does satisfy the equation with that value of \(k\).
Practice
Enter the \(y\)-coordinate, exactly, of the point where the graph of \(8x+5y=18\) meets the \(y\)-axis.
Show the solution
A point of the \(y\)-axis has \(x=0\), so \(8(0)+5y=18\), which is \(5y=18\), and \(y=\boxed{\frac{18}{5}}\). The crossing point is not required to have a whole-number coordinate, so keep the fraction in lowest terms rather than rounding it.
Practice
The graph of \(9x=20-2y\) crosses the \(x\)-axis at exactly one point. Enter the \(x\)-coordinate of that point as an exact fraction.
Show the solution
Every point on the \(x\)-axis has \(y=0\), so the equation becomes \(9x=20-2(0)=20\), and dividing by \(9\) gives \(x=\boxed{\frac{20}{9}}\). Since \(20\) is not a multiple of \(9\), the exact coordinate is that fraction, and a rounded decimal does not satisfy the equation.
Practice
A student lists \((5,9)\), \((10,6)\) and \((15,4)\) as points on the graph of \(3x+5y=60\). Plotted, the three do not fall in a straight line, so one pair is wrong. Keeping that pair's \(x\)-value, enter the correct \(y\)-value.
Show the solution
The three left sides are \(15+45=60\), \(30+30=60\) and \(45+20=65\), so \((15,4)\) is the wrong pair. At \(x=15\), \(5y=60-45=15\) and \(y=\boxed{3}\). Two points always fall in one straight line, wrong ones included, so a bad pair only shows up once a third point is plotted.
Practice
Enter the number of distinct points where the graph of \(4x-7y=0\) meets an axis.
Show the solution
Both crossings are the same point, the origin, so the answer is \(\boxed{1}\). Setting \(y=0\) gives \(4x=0\) and so \(x=0\), and setting \(x=0\) gives \(-7y=0\) and so \(y=0\), which makes both crossings \((0,0)\). Counting one crossing per axis gives \(2\) and counts the origin twice.
Practice
Enter the \(x\)-coordinate of the point where the graph of \(3(x-y)=x+38\) meets the \(x\)-axis.
Show the solution
A point on the \(x\)-axis has \(y=0\), so \(3(x-0)=x+38\), which is \(3x=x+38\). Subtracting \(x\) from both sides gives \(2x=38\), so \(x=\boxed{19}\). Setting \(x=0\) instead finds where the graph meets the \(y\)-axis, a different point. Check, \(3(19-0)=57\) and \(19+38=57\).
Practice
The graph of \(px+8y=20\) has no point in common with the \(x\)-axis, where \(p\) is a constant. Enter the \(y\)-coordinate of the point on that graph whose \(x\)-coordinate is \(-25\).
Show the solution
A point on the \(x\)-axis has \(y=0\), and the equation then reads \(px=20\). If \(p\neq 0\) this gives the point \(\left(\frac{20}{p},0\right)\) on the \(x\)-axis, so \(p=0\) and the equation is \(8y=20\). Every point of the graph has \(y=\frac{20}{8}\), so the point with \(x=-25\) has \(y=\boxed{\frac{5}{2}}\). With \(p=0\) the graph is the horizontal line \(y=\frac{5}{2}\), so the \(x\)-coordinate does not matter here.
Practice
The graph of \(8y-15x=120\) meets the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\). Enter the distance between \(A\) and \(B\).
Show the solution
Setting \(y=0\) gives \(-15x=120\), so \(x=-8\) and \(A=(-8,0)\). Setting \(x=0\) gives \(8y=120\), so \(y=15\) and \(B=(0,15)\). Then the distance is \(\sqrt{(-8-0)^2+(0-15)^2}=\sqrt{64+225}=\sqrt{289}=\boxed{17}\). Each crossing sits on an axis, so the horizontal gap is \(8\) and the vertical gap is \(15\), and \(8^2+15^2=17^2\) is why the distance comes out whole.