Three or More Variables

Some stories name three quantities and state three facts, so the system has three letters and three equations. No new method is needed for that. One substitution, or one elimination, takes the count from three letters down to two, and 5.2 and 5.3 already solve what is left. The next two problems run that once, on one system.

Problem

The system $x=4z+7$, $2x+3y+5z=255$, and $3x+y-2z=175$ has three letters, and the first equation is already solved for $x$. Write $4z+7$ in place of $x$ in both of the other two, which leaves two equations in $y$ and $z$ alone. Solve that pair and enter $z$.

Show a hint

One equation is already solved for a letter, so $4z+7$ can stand wherever $x$ stands. Do the replacement in the second equation and in the third, not in just one of them.
After the replacement the second equation collects to $3y+13z=241$ and the third to $y+10z=154$. In $y+10z=154$, $y$ has coefficient $1$.

Show the full solution

Substituting $x=4z+7$ gives $2(4z+7)+3y+5z=255$, which collects to $3y+13z=241$, and $3(4z+7)+y-2z=175$, which collects to $y+10z=154$. From $y+10z=154$, $y=154-10z$, so $3(154-10z)+13z=241$, then $462-17z=241$ and $17z=221$, so $z=\boxed{13}$. Both replacements are needed, since doing only one leaves an equation that still carries $x$ rather than a pair in $y$ and $z$.

Problem

Same system as the last problem, and $z=13$ is now known. Find $y$ and $x$, then check the triple in all three original equations and enter $x+y$.

Show a hint

One of the two reduced equations is $y+10z=154$, and with $z$ known only $y$ is left in it.
Neither reduced equation has an $x$ in it, so $x$ comes only from $x=4z+7$, the equation that was already solved for it.

Show the full solution

$y+10(13)=154$ gives $y=24$, and $x=4(13)+7=59$, so $x+y=\boxed{83}$. The triple $(59,24,13)$ passes all three originals, $59=4(13)+7$, $118+72+65=255$, and $177+24-26=175$. Two back-substitutions are needed, one for the second letter of the reduced pair and one for the letter that was replaced.

Problem

In $$3x+4y+2z=150$$ $$5x+2y-2z=108$$ $$4x+3y+6z=231$$ no letter has coefficient $1$ anywhere, so isolating one starts with a division. The $z$-terms are the cheap ones to match. Eliminate $z$, once from the first and second equations and once from the first and third. Solve the two-letter system that comes out and enter $x$.

Show a hint

The $z$-terms are $+2z$, $-2z$ and $+6z$. The first two are opposite already, and tripling the first equation matches the $6z$ in the third.
Adding the first two equations leaves $8x+6y=258$, and three times the first minus the third leaves $5x+9y=219$. Every term of $8x+6y=258$ divides by $2$.

Show the full solution

Adding the first two equations clears $z$ and gives $8x+6y=258$, or $4x+3y=129$. Three times the first equation minus the third gives $5x+9y=219$. Tripling $4x+3y=129$ gives $12x+9y=387$, and subtracting leaves $7x=168$, so $x=\boxed{24}$. Then $3y=129-96=33$ gives $y=11$, and $2z=150-72-44=34$ gives $z=17$. Check in the third equation, $96+33+102=231$.

Problem

Here is a system in three letters. $$5x+2y+3z=161$$ $$4x-2y+5z=55$$ $$3x+4y-2z=138$$ Clear $y$ from two different pairs of equations, solve the two-letter system that is left, and enter $y$.

Show a hint

The $y$-terms are $+2y$, $-2y$ and $+4y$. The first two are opposite already, and doubling the second equation makes $-4y$, the opposite of the third's $+4y$.
With those pairings the reduced pair is $9x+8z=216$ and $11x+8z=248$, both carrying $8z$, so subtracting leaves $x$ alone. The ask is $y$, which appears in neither.

Show the full solution

Adding the first two equations gives $9x+8z=216$. Adding twice the second equation to the third gives $11x+8z=248$. Subtracting leaves $2x=32$, so $x=16$, then $8z=216-144=72$ and $z=9$. Back in the first equation, $2y=161-80-27=54$, so $y=\boxed{27}$. Check, $4(16)-2(27)+5(9)=55$ and $3(16)+4(27)-2(9)=138$. The letter cleared first is not in the reduced pair, so an original equation is the only place left to find its value.

The same letter is cleared from two different pairs of rows, the first with the second and then the second with the third. What is left is two equations in $x$ and $y$, which 5.2 and 5.3 finish. Running the same step once more on those two leaves one equation in one letter.

Problem

In $$4x+3y+5z=273$$ $$7x+6y+8z=471$$ $$5x+9y-6z=77$$ one letter is cheaper to eliminate than the other two. Clear that letter from two different pairs of these equations, solve the two-letter system that is left, and enter $z$.

Show a hint

5.3 clears the letter whose coefficients meet at the smaller multiple. The two $y$-pairs used here meet at $6$ and $9$, while the cheapest $x$-pair meets at $20$ and the cheapest $z$-pair at $24$.
Doubling the first equation makes its $y$ term $6y$, matching the second, and subtracting leaves $x+2z=75$. Tripling the first makes that term $9y$, matching the third, which gives a second equation in $x$ and $z$.

Show the full solution

Doubling the first equation and subtracting the second gives $x+2z=75$. Tripling the first and subtracting the third gives $7x+21z=742$, which divides by $7$ to $x+3z=106$. Subtracting those two leaves $z=\boxed{31}$. Back-substituting gives $x=13$ and $y=22$, and $(13,22,31)$ checks in all three originals. Clearing $x$ instead would take $28x$ in one pair and $20x$ in the other, with right sides up to $1911$ instead of $819$.

A question does not have to ask for a letter. It can ask for $x+y+z$, or for the cost of one of each item. At a solution all three equations are true, so adding them adds equal numbers to equal numbers, and the result is one more true statement about that triple. Sometimes its left side is exactly the quantity the question named.

Problem

These three equations hold at once for the same $x$, $y$, and $z$. $$5x+2y+7z=238$$ $$4x+6y+3z=264$$ $$3x+4y+2z=182$$ Add all three, left side to left side and right side to right side, then enter $x+y+z$ without solving for the letters one at a time.

Show a hint

Look at the three coefficients in each letter's column. The $x$-column, the $y$-column, and the $z$-column all come to the same number here.
Adding the three equations gives $12x+12y+12z=684$, and each term there is a multiple of $12$.

Show the full solution

Adding the three left sides gives $12x+12y+12z$, and adding the three right sides gives $684$, so $12(x+y+z)=684$ and $x+y+z=\boxed{57}$. The triple is $(18,25,14)$, but reaching it takes two rounds of elimination and two back-substitutions, while one addition and one division give the sum.

Problem

$$7x+3y+5z=207$$ $$4x+8y+2z=196$$ $$5x+5y+z=145$$ The ask is $x+y+z$ and not any one letter. No letter in this system has a whole-number value, so a full solve ends in fractions for all three. Find a combination of the three equations whose left side is a multiple of $x+y+z$, then enter $x+y+z$.

Show a hint

Add the columns of all three equations first. They come to $16$, $16$ and $8$, so a plain sum is not a multiple of $x+y+z$. A minus sign on one of the three is the next thing to try.
The first plus the second minus the third leaves $6$ in every column.

Show the full solution

Adding the first two equations and subtracting the third gives $(7+4-5)x+(3+8-5)y+(5+2-1)z=207+196-145$, which is $6x+6y+6z=258$, so $x+y+z=\boxed{43}$. The triple itself is $\left(\frac{43}{4},\frac{59}{4},\frac{35}{2}\right)$, so a full solve ends in quarters for all three letters, while the combination takes one line.

This is 5.5's routine with the count raised again. Name all three quantities with units, write one equation for each stated fact, solve, then read the problem's last sentence again to see which quantity it asked for. A triangle's three angles are one such story, since the angle sum is a stated fact like any other.

Problem

A framing shop charges one fixed price to cut a mat, one to fit glass, and one to assemble a frame. One job of 3 mats cut, 2 sheets of glass fitted and 1 frame assembled came to 129 dollars. A second job of 2 mats, 4 glass sheets and 3 frames came to 237 dollars. A third of 5 mats, 1 glass sheet and 2 frames came to 166 dollars. Enter the price of fitting one sheet of glass, in dollars.

Show a hint

Three prices are unknown, so name all three with units and write one equation per job.
With $m$, $g$ and $f$ the three prices in dollars, $3m+2g+f=129$, $2m+4g+3f=237$ and $5m+g+2f=166$. The $f$-column is $1$, $3$, $2$, so scaling the first job and subtracting clears $f$ against either of the other two.

Show the full solution

With $m$, $g$ and $f$ the three prices in dollars, $3m+2g+f=129$, $2m+4g+3f=237$ and $5m+g+2f=166$. Three times the first minus the second gives $7m+2g=150$, and twice the first minus the third gives $m+3g=92$. Then $m=92-3g$, so $7(92-3g)+2g=150$, giving $644-19g=150$ and $g=\boxed{26}$. Then $m=14$ and $f=35$, and the second job checks, $28+104+105=237$.

Problem

A tea seller charges one fixed price per tin of green tea, one per tin of black tea and one per tin of herbal tea. One order of 4 green, 1 black and 3 herbal tins cost 59 dollars. A second order of 3 green, 5 black and 2 herbal cost 89 dollars. A third order of 2 green, 3 black and 4 herbal cost 77 dollars. Enter the combined cost of one tin of each, in dollars.

Show a hint

Name the three tin prices with units and write one equation per order. The ask is the three prices added together, not any one of them, so look for a combination of the three equations that lands on that sum.
Add the three green coefficients, then the three black coefficients, then the three herbal coefficients. Each of those totals is $9$, so adding the three equations gives nine of every price at once.

Show the full solution

With $g$, $b$ and $h$ the green, black and herbal tin prices in dollars, $4g+b+3h=59$, $3g+5b+2h=89$ and $2g+3b+4h=77$. The three green coefficients add to $9$, and so do the black and the herbal, so adding the three equations gives $9g+9b+9h=225$ and $g+b+h=\boxed{25}$. Solving the whole system gives $6$, $11$ and $8$ dollars, which is more work than this ask needs.

$4x+9=25$ is true for one value of $x$ and false for every other. $2(x+3)=2x+6$ is true for every value. The second is a stronger claim, and with unknown constants in it the claim is about them, not about $x$. A claim that holds at every $x$ holds at whichever values are easiest to substitute, and each gives an equation in the constants.

Problem

$a(x+6)+b(x-4)=17x+22$ is true for every value of $x$. Substitute $x=0$, then substitute $x=1$. Each substitution gives one equation in $a$ and $b$. Solve that system and enter $a$.

Show a hint

The statement holds at every $x$, so it holds at $x=0$ and at $x=1$, and each of those is an ordinary equation in $a$ and $b$.
At $x=0$ the statement reads $6a-4b=22$. At $x=1$ it reads $7a-3b=39$. That pair is 5.3's territory, and the $b$-coefficients $-4$ and $-3$ both divide $-12$.

Show the full solution

At $x=0$, $6a-4b=22$. At $x=1$, $7a-3b=39$. Tripling the first and quadrupling the second gives $18a-12b=66$ and $28a-12b=156$, and subtracting leaves $10a=90$, so $a=\boxed{9}$. Then $b=8$, and a check at $x=2$ gives, $9(8)+8(-2)=56$ and $17(2)+22=56$. The same $a$ and $b$ come out whatever two values of $x$ are used, and substituting $x=4$ or $x=-6$ makes one term zero, which pins one letter in a single line.

Problem

$a(x+2y+1)+b(3x-y+2)+c(x+y+4)=106x+19y+113$ is true for every value of $x$ and every value of $y$. Collect the left side, match the $x$ column, the $y$ column and the constants, then solve the three equations in $a$, $b$ and $c$. Enter $a$.

Show a hint

Collect the whole left side into an $x$-part, a $y$-part and a constant part before comparing anything. The $x$-part is $(a+3b+c)x$.
The three matches are $a+3b+c=106$, $2a-b+c=19$ and $a+2b+4c=113$. The first two carry the same $c$-coefficient, so subtracting one from the other clears $c$ with no scaling.

Show the full solution

Collecting gives $(a+3b+c)x+(2a-b+c)y+(a+2b+4c)$, so $a+3b+c=106$, $2a-b+c=19$ and $a+2b+4c=113$. Subtracting the second from the first clears $c$ and gives $-a+4b=87$. Subtracting the third from four times the first also clears $c$ and gives $3a+10b=311$. From $a=4b-87$, $12b-261+10b=311$, so $22b=572$ and $b=26$, which gives $a=\boxed{17}$. Then $c=11$. A statement true for every $x$ and every $y$ gives one match per column, three in all, which is exactly the number of equations needed to find $a$, $b$ and $c$.

5.1's three counts carry over. Three equations in three letters usually pin one triple, and it fails in the two ways 5.3 named. Combining them down to a false statement like $0=7$ means no triple fits all three. Reaching $0=0$ means one of the three follows from the other two, leaving two facts for three letters and infinitely many triples.

Problem

For exactly one value of $k$, the system $4x+3y+2z=158$, $5x-y+6z=143$, $9x+2y+8z=k$ has infinitely many solutions, and for every other value it has none. Enter that $k$.

Show a hint

Compare the third equation's left side with the sum of the first two left sides, column by column.
The first two left sides add to exactly the third left side, so for any triple satisfying the first two equations the value of $9x+2y+8z$ is already known. Match $k$ to it.

Show the full solution

Adding the first two equations gives $9x+2y+8z=301$, which is the third equation exactly when $k=\boxed{301}$. At that $k$ only two of the three equations are independent, and two equations in three letters have infinitely many solutions. At any other $k$, subtracting the first two equations from the third leaves $0=k-301$, which is false, so no triple works.

That closes chapter 5. A system is a set of facts about the same letters, and every method in the chapter removed letters one at a time until a single value was left, whatever the count started at. Chapter 6, Ratios, Percents and Proportion, opens with 6.1 Ratios, Simple and Subtle.

Practice these ideas

Practice

In $5x+2y+3z=281$, two of the three letters are already known, $x=23$ and $y=41$. Enter $z$.

Show the solution

Substituting gives $115+82+3z=281$, so $3z=281-197=84$ and $z=\boxed{28}$. Check, $115+82+84=281$.

Practice

Of $y=2z+5$, $3x+y+4z=149$ and $x+2y+3z=128$, the first is already solved for $y$. Substitute that expression into the other two, solve the pair in $x$ and $z$ that is left, and enter $x$.

Show the solution

Substituting gives $3x+(2z+5)+4z=149$, which collects to $3x+6z=144$, or $x+2z=48$, and $x+2(2z+5)+3z=128$, which collects to $x+7z=118$. Subtracting leaves $5z=70$, so $z=14$ and $x=48-2(14)=\boxed{20}$. Then $y=2(14)+5=33$. Check, $3(20)+33+4(14)=149$ and $20+2(33)+3(14)=128$.

Practice

A student reports $(x,y,z)=(15,9,22)$ as the solution of $2x+3y+z=79$, $4x+y+2z=113$ and $3x+5y+4z=188$. Enter how many of the three equations the triple satisfies.

Show the solution

The triple gives $30+27+22=79$, then $60+9+44=113$, then $45+45+88=178$, which misses $188$, so the triple satisfies $\boxed{2}$ of the three. The true solution is $(13,9,26)$. A triple built from two of the equations satisfies those two automatically, so it is whichever equation was left out that does the checking.

Practice

Clear $z$ from two different pairs of $2x+3y+3z=180$, $4x+2y-3z=26$ and $5x-3y+3z=147$, solve the two-letter system that comes out, and enter $y$.

Show the solution

Adding the first two equations gives $6x+5y=206$, and adding the second to the third gives $9x-y=173$. From $y=9x-173$, $6x+5(9x-173)=206$, so $51x=1071$ and $x=21$, then $y=189-173=\boxed{16}$, and $3z=180-42-48=90$ gives $z=30$. Check in the third equation, $105-48+90=147$.

Practice

In $6x+5y+7z=312$, $6x+8y+4z=348$ and $6x+4y+9z=315$, one letter has the same coefficient in all three equations. Clear that letter from two different pairs, then enter the value of that same letter.

Show the solution

Subtracting the first equation from the second gives $3y-3z=36$, so $y-z=12$, and subtracting the third from the first gives $y-2z=-3$. Subtracting those two leaves $z=15$, then $y=27$, and $6x=312-135-105=72$, so $x=\boxed{12}$. Check in the second equation, $72+216+60=348$. The reduced pair has no $x$ in it, so $x$ is found last, by substituting $y$ and $z$ into an original equation.

Practice

In $5x+2y+z=151$, $2x+5y+z=172$ and $x+y+6z=109$, enter $x+y+z$.

Show the solution

Adding the three equations gives $8x+8y+8z=432$, so $x+y+z=\boxed{54}$. The triple is $(18,25,11)$, which the question never needed. Adding works here because each letter's three coefficients total the same number.

Practice

One triple $(x,y,z)$ satisfies all three of $8x+5y+6z=387$, $8x+2y+3z=234$ and $5x+7y-4z=207$. Enter $y+z$ for that triple.

Show the solution

Subtracting the second equation from the first gives $3y+3z=153$, and dividing by 3 gives $y+z=\boxed{51}$. The full triple is $(14,31,20)$, but one subtraction is enough for $y+z$, with no use for the third equation.

Practice

The three angles of a triangle add to 180 degrees. The second angle is 21 degrees more than the first, and the first two together are 30 degrees more than the third. Enter the largest of the three angles, in degrees.

Show the solution

With $a$, $b$ and $c$ the three angles in degrees, $a+b+c=180$, $b=a+21$ and $a+b=c+30$. Substituting the third into the first gives $(c+30)+c=180$, so $c=75$ and $a+b=105$. With $b=a+21$, $2a+21=105$ gives $a=42$ and $b=63$, so the largest angle is $\boxed{75}$ degrees. Check, $42+63+75=180$. The ask is the largest angle, not the letter solved for first.

Practice

A depot stacks three sizes of box, each size the same mass. One pallet of 2 small, 3 medium and 1 large box weighs 74 kg. A second of 1 small, 2 medium and 3 large weighs 94 kg. A third of 4 small, 1 medium and 2 large weighs 103 kg. Enter the mass of one large box, in kilograms.

Show the solution

With $s$, $m$ and $l$ the three masses in kilograms, $2s+3m+l=74$, $s+2m+3l=94$ and $4s+m+2l=103$. Twice the second minus the first gives $m+5l=114$, and four times the second minus the third gives $7m+10l=273$. Then $m=114-5l$, so $798-35l+10l=273$, giving $25l=525$ and $l=\boxed{21}$. Then $m=9$ and $s=13$, and the third pallet checks, $52+9+42=103$.

Practice

Constants $a$ and $b$ make $a(x-5)+b(x+3)=24x-8$ true for every value of $x$. Enter $b$.

Show the solution

At $x=5$ the $a$-term is zero, so $8b=24(5)-8=112$ and $b=\boxed{14}$. At $x=-3$ the $b$-term is zero and $-8a=-80$ gives $a=10$. Matching coefficients instead gives $a+b=24$ and $-5a+3b=-8$, which the same pair satisfies.

Practice

Constants $a$ and $b$ make $(3a+b)x+(a-2b)=63x+7$ true for every value of $x$, not for one value only. Enter $a$.

Show the solution

Matching the $x$-parts and the constants gives $3a+b=63$ and $a-2b=7$. Doubling the first gives $6a+2b=126$, and adding $a-2b=7$ leaves $7a=133$, so $a=\boxed{19}$, with $b=6$. Check, $3(19)+6=63$ and $19-2(6)=7$.

Practice

The system $3x+4y-2z=91$, $5x-y+3z=62$ and $8x+3y+z=160$ has three equations in three letters, which is usually enough to pin down one triple. Enter the number of solutions this system has.

Show the solution

Adding the first two equations gives $8x+3y+z=153$, the same left side as the third but with $153$ on the right in place of $160$. Subtracting leaves $0=7$, false at every triple, so the count is $\boxed{0}$. The first two equations on their own have plenty of solutions, so a triple has to pass all three before it counts.

Practice

The system $x+y+z=96$, $2x-y+3z=217$, $4x+y+5z=409$ has a third equation that follows from the first two, so there are infinitely many solutions. Exactly one of them has $y=0$. Enter $x$ for that one.

Show the solution

With $y=0$, the first two equations read $x+z=96$ and $2x+3z=217$. Substituting $x=96-z$ gives $192+z=217$, so $z=25$ and $x=\boxed{71}$. Check all three, $71+0+25=96$, then $142-0+75=217$, then $284+0+125=409$. Twice the first equation plus the second gives the third, which is why the system leaves one letter free.

Practice

Only two facts are known about three letters, $5x+2y+3z=214$ and $x+4y+3z=152$, so no one letter has a single value here. Enter $x+y+z$.

Show the solution

Adding the two equations gives $6x+6y+6z=366$, so $x+y+z=\boxed{61}$. No single letter is available here, since $x$, $y$ and $z$ each take infinitely many values across the solutions, while their sum takes only this one.

Practice

Solve $2(x+y)-z=49$, $3x-(y-2z)=86$ and $x+4(y+z)=179$. Clear one letter to leave two equations in the other two, finish from there, and enter $z$.

Show the solution

Cleaning gives $2x+2y-z=49$, $3x-y+2z=86$ and $x+4y+4z=179$. Twice the second plus the first gives $8x+3z=221$, and four times the second plus the third gives $13x+12z=523$. Multiplying $8x+3z=221$ by $4$ gives $32x+12z=884$, and subtracting $13x+12z=523$ leaves $19x=361$, so $x=19$. Then $3z=221-152=69$ gives $z=\boxed{23}$, with $y=17$. Check the second original, $57-(17-46)=57+29=86$. The minus in front of the bracket reaches both terms inside.

Practice

A supplier charges fixed unit prices for valves, gaskets and couplings. One shipment of 6 valves, 2 gaskets and 3 couplings cost 111 dollars. A second shipment of 2 valves, 7 gaskets and 4 couplings cost 158 dollars. A third shipment of 3 valves, 4 gaskets and 2 couplings cost 99 dollars. Enter the cost of one valve, one gasket and one coupling together, in dollars.

Show the solution

With $v$, $g$ and $c$ the three unit prices in dollars, $6v+2g+3c=111$, $2v+7g+4c=158$ and $3v+4g+2c=99$. The first plus the second minus the third gives $5v+5g+5c=111+158-99=170$, so $v+g+c=\boxed{34}$. The separate prices are $7$, $12$ and $15$, and solving for all three takes two eliminations and two back-substitutions while the combination takes one line.

Practice

Solve the four-letter system $4w+x+y+z=132$, $w+4x+y+z=168$, $w+x+4y+z=114$ and $w+x+y+4z=153$. Enter $x$, not the whole solution.

Show the solution

Adding all four equations gives $7(w+x+y+z)=567$, so $w+x+y+z=81$. The second equation is $3x+(w+x+y+z)=168$, so $3x=87$ and $x=\boxed{29}$. The full solution is $(17,29,11,24)$. Clearing one letter at a time reaches the same answer in three rounds of elimination, while the total gets there in two lines.