QuanticaAlgebra IOpen in the courseHere is a story one letter cannot carry. Lena scored 53 points on a quiz made of 2-point and 5-point questions, and she does not know the split. With \(a\) correct 2-point answers and \(b\) correct 5-point answers, her fact reads \(2a+5b=53\). No value of \(a\) can be checked on its own. The claim involves both letters at once.
Problem
Four pairs are proposed for \(2a+5b=53\), each written with \(a\) first and \(b\) second. $$(24,1)\qquad(19,3)\qquad(9,7)\qquad(11,6)$$ Substitute both values from each pair and evaluate the left side. Enter how many of the four pairs make the claim true.
Show a hint
- Each pair gets the full check from 4.1. Replace \(a\) and \(b\) with the pair's two values, evaluate the left side, and compare it with \(53\).
- The last pair gives \(2(11)+5(6)=22+30\). Settle whether that is \(53\), then run the other three the same way.
Show the full solution
The first three pairs pass, \(48+5=53\), \(38+15=53\), and \(18+35=53\), while \((11,6)\) gives \(22+30=52\). So \(\boxed{3}\) of the four pairs make the claim true. Three different pairs already satisfy the same equation, and nothing in it stops more.
Problem
The pair \(x=12,\ y=2\) solves \(4x+9y=66\), since \(48+18=66\). Now swap the two values. Substitute \(x=2,\ y=12\) and enter the value the left side takes.
Show a hint
- A pair says which value belongs to which letter, so the swapped version is a different pair making a different substitution.
- With \(x=2\) and \(y=12\), the first term is \(4(2)=8\) and the second is \(9(12)\).
Show the full solution
With \(x=2\) and \(y=12\), the left side is \(4(2)+9(12)=8+108=\boxed{116}\). That is not 66, so the swapped pair fails. A pair carries its order, and \((12,2)\) and \((2,12)\) are different candidates.
Checking given pairs is one skill. Producing your own is the next. Choose a value for one variable and the equation collapses to one variable, which 4.2 solves. In Lena's equation, \(a=14\) gives \(28+5b=53\), so \(5b=25\) and \(b=5\), the pair \((14,5)\). The story allows only whole counts, but the equation is its own object and takes any number.
Problem
The equation \(x+2y=49\) has a solution pair with \(x=17\). Substitute that value, solve the equation that remains, and enter \(y\).
Show a hint
- Substituting the given \(x\) leaves an equation in \(y\) alone, exactly the kind 4.2 solves.
- After the substitution the claim reads \(17+2y=49\). Solve \(2y=32\).
Show the full solution
Substituting \(x=17\) gives \(17+2y=49\), so \(2y=32\) and \(y=\boxed{16}\). The pair \((17,16)\) checks, \(17+32=49\).
Problem
The equation \(6x+y=42\) has a solution pair with \(x=10\). Substitute the given value, solve for the partner, and enter \(y\).
Show a hint
- Replace \(x\) with \(10\) and compute \(6x\) first. It is fine if that value comes out larger than \(42\).
- Substituting gives \(60+y=42\). Since \(60\) is already more than \(42\), \(y\) must be negative.
Show the full solution
Substituting \(x=10\) gives \(60+y=42\), so \(y=42-60=\boxed{-18}\). The check passes, \(60+(-18)=42\). A solution pair can include a negative number.
Problem
The equation \(3x+10y=58\) has a solution pair with \(x=18\). Find the partner value and enter \(y\) as a fraction in lowest terms.
Show a hint
- The collapsed equation is 4.2 work, and nothing promises a whole-number answer.
- Substituting gives \(54+10y=58\). Solve \(10y=4\), then reduce.
Show the full solution
Substituting \(x=18\) gives \(54+10y=58\), so \(10y=4\) and \(y=\frac{4}{10}=\boxed{\tfrac{2}{5}}\). Nothing restricts pairs to whole numbers. Every choice of \(x\), whole or not, gets a partner.
Two secret numbers satisfy \(x+y=101\). Finding them looks hopeless, since pairs passing that test can be listed all day and every one of them keeps the secret. Suppose a second fact about the same two numbers turns up. It is worth asking what one more fact could possibly change.
Problem
All four of these pairs satisfy \(x+y=101\). $$(98,3)\qquad(93,8)\qquad(79,22)\qquad(70,31)$$ Now the second fact arrives. The same two numbers also satisfy \(x-y=39\). Enter how many of the four pairs satisfy both facts.
Show a hint
- Each pair already passes the first fact, so only the new claim needs testing.
- The first pair gives \(98-3=95\), not \(39\). Keep going down the list.
Show the full solution
The differences are \(98-3=95\), \(93-8=85\), \(79-22=57\), and \(70-31=39\). Only \((70,31)\) gives \(39\), so \(\boxed{1}\) pair satisfies both facts. The first fact allowed an endless list, and the second fact cut it to one pair.
Lena knew one more fact all along. The quiz had 13 questions, so \(a+b=13\) must also hold. Of every pair passing \(2a+5b=53\), only \((4,9)\) passes both, since \(8+45=53\) and \(4+9=13\). She got four 2-point questions and nine 5-point questions. Checking a pair against a system means checking it against every equation in it.
Problem
For the system \(2u+v=69\) and \(u+v=37\), test the pairs \((30,9)\), \((32,5)\), and \((26,11)\). One satisfies only the first equation, one satisfies only the second, and one satisfies both. Enter \(u\) for the solution of the system.
Show a hint
- A solution of a system passes every equation in it. A pair that fails one check is not a solution, even if it passes every other.
- \((30,9)\) makes the first equation true, since \(60+9=69\). Check it in the second before deciding anything.
Show the full solution
\((30,9)\) gives \(60+9=69\) but \(30+9=39\), not \(37\). \((26,11)\) gives \(26+11=37\) but \(52+11=63\), not \(69\). \((32,5)\) gives \(64+5=69\) and \(32+5=37\), so it passes both and \(u=\boxed{32}\). Each rejected pair passed exactly one equation, and one pass is worth nothing in a system.
Problem
Both \((13,7)\) and \((29,3)\) satisfy \(x+4y=41\). Only one of them also satisfies \(2x-y=19\). Enter \(x+y\) for the solution of the system.
Show a hint
- Passing the first equation is not enough. Run each pair through the second and keep the one that passes.
- For \((29,3)\), the left side of the second equation is \(58-3\). Compare that with \(19\).
Show the full solution
For \((29,3)\), \(2x-y=58-3=55\), not \(19\). For \((13,7)\), \(2x-y=26-7=19\), and \(13+28=41\) confirms the first. The solution is \((13,7)\), so \(x+y=13+7=\boxed{20}\).
How is that single pair found when no candidates are given? For now, by search. Take the friendlier equation, list whole-number pairs that satisfy it, and test each one in the other equation until a pair passes both. This works when the numbers are kind, and it fails the moment they are not. Lessons 5.2 and 5.3 build methods that always work.
Problem
Two numbers satisfy \(x+y=23\) and \(5x-y=25\). List whole-number pairs from the first equation and test them in the second until one passes. Enter \(x\).
Show a hint
- Every candidate from the first fact has the form \(x\) and \(23-x\), so each choice of \(x\) names a full pair to test.
- If a test comes out below \(25\), raise \(x\) and try again. The value of \(5x-y\) climbs quickly as \(x\) grows, so the search stays short.
Show the full solution
Candidates from the first equation include \((7,16)\), \((8,15)\), and \((9,14)\). Testing \((8,15)\) in the second, \(5(8)-15=40-15=25\), so both facts hold and \(x=\boxed{8}\). Raising \(x\) by \(1\) raises \(5x-y\) by \(6\), so the second equation is hit exactly once.
Problem
The system \(m+n=67\) and \(m+n=73\) makes two claims about the same two numbers. Take any pair satisfying the first equation, test it in the second, and enter the number of solutions of the system.
Show a hint
- Read the two claims side by side. Both are about the sum of the same two numbers.
- A pair with \(m+n=67\) makes the second left side \(67\), and \(67\) is not \(73\).
Show the full solution
Any pair passing the first equation has \(m+n=67\), so in the second equation its left side is \(67\), never \(73\). No pair passes both, and the system has \(\boxed{0}\) solutions. Two claims about the same numbers can simply contradict each other. Chapter 7 will draw what that looks like.
Problem
Two numbers satisfy \(x+y=6\) and \(x-y=14\). Search pairs from the first fact, and do not stop at positive candidates. Enter \(y\).
Show a hint
- The difference \(14\) is larger than the sum \(6\). Two positive numbers always differ by less than their sum, so one member must be negative.
- Pairs summing to 6 include ones with a negative member, like \(9\) and \(-3\).
Show the full solution
Pairs from the first fact run \((7,-1)\), \((8,-2)\), \((9,-3)\), \((10,-4)\). Testing \((10,-4)\) in the second, \(10-(-4)=14\), so both facts hold and \(y=\boxed{-4}\). Positive pairs summing to \(6\) differ by less than \(6\), so the pair that works has to include a negative member.
A system is two claims sharing one pair. Testing confirms a pair, and searching finds one while the numbers stay kind, but they will not stay kind. Lesson 5.2, Substitution, turns two equations in two variables into one equation in one variable, chapter 4's home ground.
Practice these ideas
Practice
Substitute the pair \(x=11,\ y=2\) into \(8x+3y=94\) and evaluate the left side. It matches the right side. Enter the number both sides equal.
Show the solution
The left side is \(8(11)+3(2)=88+6=\boxed{94}\). One check settles a pair, both values go in together and the two sides are compared once.
Practice
The pair \((7,6)\) is claimed to solve \(9x+4y=91\). Substitute \(x=7,\ y=6\) and enter the value the left side actually takes.
Show the solution
\(9(7)+4(6)=63+24=\boxed{87}\). That is not \(91\), so \((7,6)\) is not a solution. A check either matches the right side or it does not, and this one misses by \(4\).
Practice
The equation \(7x+y=86\) has a solution pair with \(x=11\). Substitute that value and solve for the letter that remains. Enter \(y\).
Show the solution
Substituting \(x=11\) gives \(77+y=86\), so \(y=\boxed{9}\). Check, \(7(11)+9=77+9=86\), and the pair \((11,9)\) passes.
Practice
The equation \(x-2y=11\) has a solution pair with \(y=12\). Enter \(x\).
Show the solution
Substituting \(y=12\) gives \(x-24=11\), so \(x=\boxed{35}\). Check, \(35-2(12)=35-24=11\). Either member of the pair can be the given one.
Practice
Check the pairs \((6,0)\), \((4,6)\), \((3,8)\), \((5,2)\) against \(3x+y=18\), substituting \(x\) first and \(y\) second. Enter how many of the four are solutions.
Show the solution
The left sides are \(18+0=18\), \(12+6=18\), \(9+8=17\), \(15+2=17\), so \(\boxed{2}\) of the pairs are solutions. Zero is a legal value for a member of a pair, and \((6,0)\) passes like any other.
Practice
Substitute the pair \((24,9)\) into \(x+9y=4x+y\), evaluating each side separately. Both sides come out to the same number. Enter it.
Show the solution
The left side is \(24+81=105\) and the right side is \(96+9=105\), so both sides equal \(\boxed{105}\). The pair checks against the whole claim at once, whichever side each letter appears on.
Practice
The equation \(8x+y=50\) has a solution pair with \(x=9\). Enter \(y\).
Show the solution
Substituting \(x=9\) gives \(72+y=50\), so \(y=50-72=\boxed{-22}\). Check, \(8(9)+(-22)=72-22=50\). The pair has a negative member, and the check passes all the same.
Practice
The equation \(8x+6y=85\) has a solution pair with \(x=8\). Enter \(y\) as a fraction in lowest terms.
Show the solution
Substituting \(x=8\) gives \(64+6y=85\), so \(6y=21\) and \(y=\frac{21}{6}=\boxed{\frac{7}{2}}\). Check, \(64+6\cdot\frac{7}{2}=64+21=85\). Nothing in the equation forces \(y\) to be a whole number.
Practice
The equation \(4y-3x=21\) has a solution pair with \(x=9\). Enter \(y\).
Show the solution
Substituting \(x=9\) gives \(4y-27=21\), so \(4y=48\) and \(y=\boxed{12}\). Check, \(4(12)-3(9)=48-27=21\).
Practice
Every pair solving \(x+5y=88\) also solves \(3x+15y=264\), since the second equation is the first with both sides multiplied by \(3\). The second fact adds nothing new. Enter \(y\) for the pair with \(x=13\).
Show the solution
Using the first equation, \(13+5y=88\), so \(5y=75\) and \(y=\boxed{15}\). Check in the second, \(3(13)+15(15)=39+225=264\). A second equation only pins the pair when it says something the first did not.
Practice
All four of \((35,11)\), \((37,7)\), \((39,3)\), \((28,25)\) satisfy \(2x+y=81\). Exactly one also satisfies \(3x-y=104\). Enter \(x\) for that pair.
Show the solution
Testing the second equation, \(105-11=94\), \(111-7=104\), \(117-3=114\), \(84-25=59\). Only \((37,7)\) passes both, so \(x=\boxed{37}\). A solution of a system must satisfy every equation, and failing even one rules a pair out.
Practice
Two numbers add to \(87\) and differ by \(33\). Enter the larger number.
Show the solution
The facts are \(x+y=87\) and \(x-y=33\). Searching pairs that sum to \(87\), the pair \((60,27)\) gives \(60-27=33\), so the larger number is \(\boxed{60}\). Check, \(60+27=87\).
Practice
Two numbers satisfy \(x+y=29\) and \(4x+y=71\). Test whole-number pairs from the first equation in the second until one passes, then enter the product \(xy\).
Show the solution
The pair \((14,15)\) sums to \(29\), and \(4(14)+15=56+15=71\) passes the second equation. So \(xy=14\cdot 15=\boxed{210}\). No other whole pair works, since raising \(x\) by \(1\) raises \(4x+y\) by \(3\), so \(71\) is hit exactly once.
Practice
For the system \(x+y=102\) and \(x+y=89\), any pair satisfying the first equation misses the second by one fixed amount. Enter that amount.
Show the solution
Any pair with \(x+y=102\) makes the second equation read \(102=89\), which misses by \(102-89=\boxed{13}\). The two claims conflict for every pair at once, so this system has no solution.
Practice
The system \(x+y=112\) and \(x-y=112\) looks like the impossible kind, but the left sides differ, and it has exactly one solution. Enter \(x\).
Show the solution
If adding \(y\) and subtracting \(y\) leave the same value, then \(y=0\), and the first equation gives \(x=\boxed{112}\). Check, \(112+0=112\) and \(112-0=112\). Zero is a legal member of a pair, so the two claims agree instead of conflicting.
Practice
Two numbers satisfy \(m+n=5\) and \(m-n=17\). Search pairs from the first fact, and let the second number go negative. Enter \(n\).
Show the solution
The pair \((11,-6)\) gives \(11+(-6)=5\) and \(11-(-6)=11+6=17\), so \(n=\boxed{-6}\). A search that stops at positive candidates never finds this pair, since positive pairs summing to \(5\) differ by at most \(5\).
Practice
For one particular number \(k\), the pair \(x=16,\ y=-9\) is the solution of the system \(x+y=7\) and \(2x+5y=k\). Enter \(k\).
Show the solution
The first equation checks, \(16+(-9)=7\). Substituting the pair into the second, \(k=2(16)+5(-9)=32-45=\boxed{-13}\). Keep the parentheses, \(5(-9)\) is \(-45\), and the term joins the sum with its sign.