# Solving two equations at once: where lines meet

> Ten heads and 28 legs: how many chickens and rabbits? Each clue is a line of possibilities, and elimination pivots a line about the point where they cross.

Series: Algebra (https://www.monocurl.com/monocurl-essays/algebra/)
Canonical: https://www.monocurl.com/monocurl-essays/algebra/systems-of-equations/
Keywords: systems of equations, simultaneous equations, elimination method, substitution method, where two lines intersect, no solution or infinitely many solutions, chickens and rabbits problem

[Figure 1, animation: Ten animals, chickens and rabbits. Starting from ten chickens with 20 legs, each chicken swapped for a rabbit moves one step along the line of ten-head combinations and adds two legs. After four swaps the count reaches 28 legs, at 6 chickens and 4 rabbits, which is where the line of 28-leg combinations crosses.]

A farmer keeps chickens and rabbits in one pen. Looking over the fence you count 10 heads, and looking under it you count 28 legs. How many chickens are there, and how many rabbits?

Neither count answers the question by itself. Ten heads could be ten chickens, or ten rabbits, or anything in between. Twenty-eight legs could be seven rabbits, or fourteen chickens, or several other mixtures. The answer comes only from holding both facts at once. This puzzle is at least 1500 years old (it appears in the Chinese text *Sunzi Suanjing* with 35 heads and 94 feet), and the idea that solves it is the idea behind every system of equations.

**Pause and try.** Before reading on, solve the puzzle by any method you like. Then try to explain why there is only one answer.

**Answer.** There are 6 chickens and 4 rabbits: $6 + 4 = 10$ heads and $6 \times 2 + 4 \times 4 = 12 + 16 = 28$ legs. The rest of this lesson gives three ways to find this and one reason it is the only possibility.

## Handing out the legs

Here is a way to solve it that needs no algebra. Every animal, chicken or rabbit, has at least two legs. So begin by giving each of the ten heads two legs. That uses 20 legs, and 8 are left over. The leftover legs can only belong to rabbits, and each rabbit needs exactly two more. Eight legs in pairs is four pairs, so there are 4 rabbits, and the other 6 animals are chickens.

[Figure 2, steps: Solving the chickens and rabbits puzzle by handing out legs. Ten heads each receive two legs, which uses 20 of the 28. The remaining 8 legs go out in pairs to 4 animals, which are therefore the rabbits. The other 6 are chickens.]

  - step: Ten heads. Each belongs to a chicken or a rabbit, and there are 28 legs to hand out.
  - step: Every animal has at least two legs. Give each head two: that uses 20.
  - step: There are 8 legs left. Only rabbits take more, two more each, so the 8 go out in pairs to 4 animals.
  - step: Those 4 animals are the rabbits. The other 6 are chickens.
That argument is quick, and it looks like a trick that depends on legs coming in twos and fours. It is actually a general method in disguise, and the way to see that is to write the puzzle in symbols and then draw it.

## Each clue is a line of possibilities

Let $c$ be the number of chickens and $r$ the number of rabbits. The two counts say

$$
c + r = 10 \qquad\text{and}\qquad 2c + 4r = 28.
$$

Take the first equation alone. It has many solutions: $(c, r) = (10, 0), (9, 1), (8, 2)$, and so on down to $(0, 10)$. Plot each as a point, with chickens measured across and rabbits measured up, and the eleven points fall on a straight line. The second equation alone has its own list: $(14, 0), (12, 1), (10, 2), \dots, (0, 7)$, which falls on a different straight line. Step through the figure.

[Figure 3, steps: The combinations of chickens and rabbits allowed by each clue. The eleven combinations with 10 heads lie on one line, the eight combinations with 28 legs lie on another, and exactly one point, 6 chickens and 4 rabbits, is on both.]

  - step: Every way to have 10 heads, from 10 chickens and no rabbits to 10 rabbits and no chickens.
  - step: The eleven points lie on a straight line. Each step trades one chicken for one rabbit.
  - step: Every way to have 28 legs. Each step trades two chickens for one rabbit.
  - step: Only one combination is on both lists: 6 chickens and 4 rabbits.
The picture explains why the answer is unique. Each clue confines the answer to a line, and two different straight lines that cross do so at a single point.

We should check that these sets of points really are straight lines. [The lesson on slope](@algebra/slope-and-lines) proved that the graph of $y = mx + b$ is a straight line with slope $m$ crossing the vertical axis at height $b$. Our equations can be put in that form by solving for $r$:

$$
c + r = 10 \;\text{ becomes }\; r = -c + 10, \qquad 2c + 4r = 28 \;\text{ becomes }\; r = -\tfrac12 c + 7.
$$

The slopes $-1$ and $-\frac12$ are the trades in the figure: one rabbit gained per chicken lost on the first line, half a rabbit per chicken on the second.

**Definition.** A **linear equation** in the unknowns $x$ and $y$ is an equation of the form $ax + by = c$, where $a$, $b$, $c$ are given real numbers and $a$ and $b$ are not both zero. A **solution** is a pair $(x, y)$ that makes it true. A **system** of two linear equations is a pair of such equations, and a **solution of the system** is a pair $(x, y)$ that makes both true at once.

The solutions of one linear equation always form a straight line. If $b \ne 0$, dividing by $b$ gives $y = -\frac{a}{b}x + \frac{c}{b}$, a line of slope $-\frac{a}{b}$. If $b = 0$, then $a \ne 0$ and the equation says $x = \frac{c}{a}$, a vertical line. So a solution of a system is a point that lies on two lines.

## Substitution: walk along one line until the other is satisfied

Reading the crossing off a drawing works when the answer is a pair of small whole numbers. We want a method that works by calculation.

The first equation says $r = 10 - c$: once you choose $c$, the line tells you $r$. So every point on the first line has the form $(c, 10 - c)$, and we can ask which of these points satisfies the second equation. Replace $r$ by $10 - c$ there:

$$
2c + 4(10 - c) = 28, \quad\text{so}\quad 40 - 2c = 28, \quad\text{so}\quad c = 6.
$$

Then $r = 10 - 6 = 4$. This is the **substitution method**, and the animation at the top of the page is a picture of it. The marker walks along the ten-heads line, the leg count $2c + 4(10 - c) = 40 - 2c$ changes by 2 at each step, and we stop when it reads 28.

## Elimination: a third line through the same point

The leg-handing argument did something different. It never solved for one unknown in terms of the other. In symbols, "give every head two legs" is the first equation multiplied by 2:

$$
2c + 2r = 20.
$$

"See what is left over" is subtracting that from the second equation, left side from left side and right side from right side:

$$
(2c + 4r) - (2c + 2r) = 28 - 20, \quad\text{that is,}\quad 2r = 8.
$$

The chickens have disappeared from the equation, which is why this is called **elimination**. It is not obvious that one equation may be subtracted from another, so the step needs a reason.

At a solution, both equations are true statements about numbers. If $2c + 4r$ really is 28 and $2c + 2r$ really is 20, then their difference really is 8. So any solution of the system must also satisfy the new equation $2r = 8$. Put geometrically: the new equation is a new line, and it passes through the crossing point of the old two.

Nothing forced us to subtract exactly two copies of the first equation. Subtract $t$ copies, for any number $t$, and you get

$$
(2 - t)c + (4 - t)r = 28 - 10t,
$$

a whole family of lines, every one of which must pass through the crossing.

**Pause and try.** Take $t = 3$. Write down the combined equation and check that $(6, 4)$ satisfies it. Then find the value of $t$ that eliminates $r$ instead of $c$, and solve the resulting equation.

**Answer.** With $t = 3$: $-c + r = -2$, and indeed $-6 + 4 = -2$. To eliminate $r$ we need $4 - t = 0$, so $t = 4$, which gives $-2c = -12$ and $c = 6$.

[Figure 4, animation: The legs equation with t copies of the heads equation subtracted from it, as t runs from 0 to 4. The combined line pivots about the crossing point (6, 4). At t = 2 the chickens are eliminated and the line is horizontal, r = 4. At t = 4 the rabbits are eliminated and the line is vertical, c = 6.]

As $t$ changes, the combined line swings around the crossing point like a clock hand. Among all those lines two are especially easy to read, the horizontal one and the vertical one, because a horizontal line states the value of $r$ and a vertical line states the value of $c$. Elimination is the art of choosing the multiple that swings the line flat or upright.

We argued that every solution of the original pair satisfies the combined equation. For the method to be safe we need the reverse as well: replacing an equation by a combination must not let in new, false solutions.

**Theorem.** Let $E_1$ be the equation $a_1x + b_1y = c_1$ and $E_2$ the equation $a_2x + b_2y = c_2$, and let $k$ be any real number. Let $E_3$ be the equation obtained by adding $k$ times $E_1$ to $E_2$:

$$
(a_2 + ka_1)x + (b_2 + kb_1)y = c_2 + kc_1.
$$

Then a pair $(x, y)$ satisfies both $E_1$ and $E_2$ if and only if it satisfies both $E_1$ and $E_3$.

**Proof.** Suppose $(x, y)$ satisfies $E_1$ and $E_2$. Then $a_1x + b_1y$ is the number $c_1$ and $a_2x + b_2y$ is the number $c_2$, so

$$
(a_2 + ka_1)x + (b_2 + kb_1)y = (a_2x + b_2y) + k(a_1x + b_1y) = c_2 + kc_1,
$$

where the first step regroups the terms using the distributive law. So $(x, y)$ satisfies $E_3$.

Conversely, suppose $(x, y)$ satisfies $E_1$ and $E_3$. Subtract $k(a_1x + b_1y) = kc_1$ from $E_3$ in the same way. What remains is $a_2x + b_2y = c_2$, which is $E_2$.

Two things about this statement matter. It keeps $E_1$: the pair $\{E_1, E_3\}$ has the same solutions as the pair $\{E_1, E_2\}$, but $E_3$ alone is only one line and has many more. And the step can be undone, which is what the second half of the proof uses. A second safe step is to multiply both sides of one equation by a number that is not zero, since dividing by that number undoes it.

**Key idea.** Each linear equation is a line. Adding a multiple of one equation to another gives a new line through the same crossing point. Elimination chooses the multiple that makes the new line horizontal or vertical, so that it reads off one coordinate.

## Worked examples

**Worked example.** **Substitution when both equations are solved for $y$.** The slope lesson compared two phone plans, $y = 3x + 4$ and $y = 2x + 12$, where $x$ is gigabytes used and $y$ is the bill in dollars. Find the usage at which they cost the same.

Both equations give $y$, so substitute the first expression for $y$ into the second: $3x + 4 = 2x + 12$. Subtracting $2x$ and 4 from both sides gives $x = 8$. Then $y = 3 \times 8 + 4 = 28$.

Check in the other equation: $2 \times 8 + 12 = 28$. The plans cost the same, \$28, at 8 gigabytes.

**Worked example.** **Elimination when no coefficient matches.** Solve $3x + 2y = 16$ and $5x - 3y = 14$.

Neither unknown can be eliminated by a plain subtraction. We want the $y$ terms to cancel, and their coefficients are $2$ and $-3$, so we scale the equations until both carry a 6. Multiply the first by 3 and the second by 2:

$$
9x + 6y = 48, \qquad 10x - 6y = 28.
$$

Now add them. The $y$ terms cancel and $19x = 76$, so $x = 4$. Put this into the first original equation: $12 + 2y = 16$, so $y = 2$.

Check in the equation we did not use for the last step: $5 \times 4 - 3 \times 2 = 20 - 6 = 14$.

[Figure 5, image: The lines 3x + 2y = 16 and 5x - 3y = 14 cross at (4, 2). The two equations produced by elimination, 19x = 76 and 19y = 38, are the vertical and horizontal lines through the same point.]

**Worked example.** **A word problem.** A theater sells adult tickets for \$12 and child tickets for \$7. One evening it sells 50 tickets for a total of \$465. How many of each?

Name the unknowns: $a$ adult tickets and $k$ child tickets. Counting tickets gives $a + k = 50$. Counting dollars gives $12a + 7k = 465$. This has the same shape as the farm puzzle, so use the same idea: charge everyone the child price first. Seven times the first equation is $7a + 7k = 350$. Subtracting it from the second leaves $5a = 115$, so $a = 23$, and then $k = 50 - 23 = 27$.

Check: $23 \times 12 + 27 \times 7 = 276 + 189 = 465$.

## When there is no crossing, or too many

Two lines do not always cross at one point. Suppose the pen holds chickens and ducks, both two-legged, and you count 10 heads and 28 legs.

**Pause and try.** Write the two equations for $c$ chickens and $d$ ducks. Try to eliminate $c$. What happens, and what does it say about the count?

**Answer.** The equations are $c + d = 10$ and $2c + 2d = 28$. Subtracting twice the first from the second eliminates $c$ and $d$ together and leaves $0 = 8$, which is false. No pair $(c, d)$ can satisfy both equations: ten two-legged birds have 20 legs, never 28, so someone miscounted.

In the picture, $2c + 2d = 28$ is the line $c + d = 14$, which has the same slope as $c + d = 10$ and sits above it. The lines are parallel and never meet. If the leg count had been 20, the second equation would be $2c + 2d = 20$, which is the first equation doubled. Then the two lines coincide and every point of the line is a solution: the leg count tells you nothing that the head count had not.

[Figure 6, image: The three possibilities for two linear equations. Lines with different slopes cross at exactly one point. Parallel lines never meet, so there is no solution. Two equations for the same line share every point, so there are infinitely many solutions.]

These are the only three things that can happen, and one number computed from the coefficients tells them apart.

**Theorem.** Consider the system $a_1x + b_1y = c_1$, $a_2x + b_2y = c_2$, in which $a_1, b_1$ are not both zero and $a_2, b_2$ are not both zero. Let $D = a_1b_2 - a_2b_1$.

- If $D \ne 0$, the system has exactly one solution, namely $x = \dfrac{c_1b_2 - c_2b_1}{D}$, $y = \dfrac{a_1c_2 - a_2c_1}{D}$.
- If $D = 0$, the system has either no solution or infinitely many.

**Proof.** *The case $D \ne 0$.* Suppose $(x, y)$ is a solution. Multiply the first equation by $b_2$ and the second by $b_1$, and subtract; the $y$ terms cancel and $(a_1b_2 - a_2b_1)x = c_1b_2 - c_2b_1$. Multiply the first by $a_2$ and the second by $a_1$, and subtract the first from the second; the $x$ terms cancel and $(a_1b_2 - a_2b_1)y = a_1c_2 - a_2c_1$. Some of the multipliers $a_1, a_2, b_1, b_2$ may be zero, and that does no harm here: we claim only that the two new equations follow from the old ones, which is true for any multipliers. Dividing by $D$, which is not zero, shows that a solution can only be the pair in the statement. So there is at most one.

That pair really is a solution. Substituting it into the left side of the first equation gives

$$
\frac{a_1(c_1b_2 - c_2b_1) + b_1(a_1c_2 - a_2c_1)}{D} = \frac{c_1(a_1b_2 - a_2b_1)}{D} = c_1,
$$

because the two terms containing $c_2$ cancel. The second equation is checked the same way.

*The case $D = 0$.* First, the second pair of coefficients is a multiple of the first: there is a number $k$ with $a_2 = ka_1$ and $b_2 = kb_1$. Indeed, if $a_1 \ne 0$, let $k = a_2/a_1$; then $D = 0$ says $a_1b_2 = a_2b_1$, so $b_2 = kb_1$. If $a_1 = 0$, then $b_1 \ne 0$ and $D = 0$ says $a_2b_1 = 0$, so $a_2 = 0$, and $k = b_2/b_1$ works. Now subtract $k$ times the first equation from the second, which is adding $-k$ times it. The left side becomes $(a_2 - ka_1)x + (b_2 - kb_1)y = 0$, so by the previous theorem the system has the same solutions as the first equation together with $0 = c_2 - kc_1$. If $c_2 \ne kc_1$, that statement is false and there are no solutions. If $c_2 = kc_1$, it is true and imposes nothing, so the solutions are all the points of the first line, of which there are infinitely many.

When $b_1$ and $b_2$ are not zero, the two slopes are $-\frac{a_1}{b_1}$ and $-\frac{a_2}{b_2}$, and they are equal exactly when $a_1b_2 = a_2b_1$. So $D = 0$ says that the lines have the same direction. The number $D$ is called the **determinant** of the system, and it has a geometric meaning of its own, as [the factor by which a matrix scales areas](@linear-algebra/determinant). For the farm, $D = 1 \times 4 - 2 \times 1 = 2$, and the formulas give $c = \frac{10 \times 4 - 28 \times 1}{2} = 6$ and $r = \frac{1 \times 28 - 2 \times 10}{2} = 4$.

## A solution the farm cannot use

The theorem is about real numbers. A puzzle about animals wants whole numbers that are not negative, and the algebra does not know that. Change the leg count to 27. Then $D$ is still 2, so there is exactly one solution: twice the heads equation subtracted from $2c + 4r = 27$ leaves $2r = 7$, so $r = 3.5$ and $c = 6.5$.

[Figure 7, image: Ten heads and 27 legs. The lines c + r = 10 and 2c + 4r = 27 cross at (6.5, 3.5), which falls between the grid points. The system has a solution, but no whole number of chickens and rabbits matches the count.]

Half a rabbit is not an answer, so the count of 27 must be wrong. (You could have known in advance: $2c + 4r$ is even for whole numbers $c$ and $r$.) In the same way, 10 heads and 44 legs gives $r = 12$ and $c = -2$, a crossing point outside the part of the plane where the puzzle lives. After solving a system that came from a situation, check the solution against the situation.

## Common mistakes

**Common mistake.** **Subtracting only part of an equation.** To eliminate $x$ from $2x + 3y = 13$ and $2x - y = 1$, a student subtracts and writes $2y = 12$.

The $y$ terms were combined as $3y - y$, but subtracting the second equation means subtracting $-y$, and $3y - (-y) = 4y$. The correct result is $4y = 12$, so $y = 3$ and $x = 2$. A check catches this at once: the student's $y = 6$ gives $x = -\frac52$ from the first equation, and then $2x - y = -11$, not 1. Subtract every term, and when signs are mixed prefer to multiply one equation by $-1$ and add.

**Common mistake.** **Substituting back into the equation you started from.** From $c + r = 10$ a student gets $r = 10 - c$, then substitutes into $c + r = 10$ again and finds $c + (10 - c) = 10$, that is, $10 = 10$, and concludes that every $c$ works.

The statement $10 = 10$ is true, and it says only that points on the first line are on the first line. The expression $r = 10 - c$ already contains everything the first equation knows. It has to be substituted into the *other* equation, which is the one carrying new information.

## Stepping back

One equation with two unknowns leaves a line of possibilities. A second equation leaves another line, and what both allow is where the lines meet. The difficulty was that a crossing point is easy to see and awkward to compute, and the idea that removed it was that equations may be combined: any multiple of one added to the other gives a new line through the same point, and we are free to choose the one that is horizontal or vertical.

That idea scales up. With three unknowns each equation is a plane in space, and with a thousand unknowns there is no picture at all, but adding multiples of equations to one another to eliminate unknowns one at a time still works. Under the name Gaussian elimination it is how computers solve the large systems that arise in engineering and statistics, and [the lesson on linear systems](@linear-algebra/linear-systems-and-inverses) explains the same three cases, one solution, none or infinitely many, by treating the two left sides as a single transformation of the plane.

**Check yourself.** The system $x + 2y = 7$, $3x + 6y = 21$ has how many solutions?
Answer: Infinitely many The second equation is three times the first, so both describe the same line, and every point of it, such as $(7, 0)$, $(5, 1)$ and $(1, 3)$, is a solution.

**Recap.** - The solutions of one linear equation $ax + by = c$ form a straight line. A solution of a system of two is a point on both lines.
- Substitution: solve one equation for one unknown and substitute into the other equation.
- Elimination: add a multiple of one equation to the other so that one unknown cancels. The combined equation is a line through the same crossing point, chosen to be horizontal or vertical.
- These steps do not change the set of solutions, because each can be undone.
- With $D = a_1b_2 - a_2b_1$: if $D \ne 0$ there is exactly one solution; if $D = 0$ the lines are parallel (no solution) or identical (infinitely many).
- Check a solution in both original equations, and against the situation it came from.

## Problems

**Problem 1.** Solve by substitution: $y = 2x - 1$ and $3x + 2y = 12$.

**Solution.** The first equation gives $y$ in terms of $x$, so substitute it into the second: $3x + 2(2x - 1) = 12$, that is, $7x - 2 = 12$, so $x = 2$. Then $y = 2 \times 2 - 1 = 3$.

Check in the second equation: $3 \times 2 + 2 \times 3 = 12$. The solution is $(2, 3)$.

**Problem 2.** Solve by elimination: $x + y = 9$ and $x - y = 3$. Then describe the two lines and their crossing point.

**Solution.** Adding the equations cancels $y$: $2x = 12$, so $x = 6$. Subtracting the second from the first cancels $x$: $2y = 6$, so $y = 3$.

Check: $6 + 3 = 9$ and $6 - 3 = 3$. The first line is $y = -x + 9$, falling with slope $-1$, and the second is $y = x - 3$, rising with slope 1. They cross at $(6, 3)$.

**Problem 3.** A parking lot holds motorcycles and cars: 30 vehicles and 96 wheels in all. Solve it twice, once by the "hand out the wheels" argument in words, and once by writing equations and eliminating.

**Solution.** *In words.* Give every vehicle two wheels: that uses 60. The remaining $96 - 60 = 36$ wheels belong to cars, two more for each, so there are 18 cars and $30 - 18 = 12$ motorcycles.

*With equations.* Let $m$ be the number of motorcycles and $k$ the number of cars: $m + k = 30$ and $2m + 4k = 96$. Subtracting twice the first equation from the second gives $2k = 36$, so $k = 18$ and $m = 12$.

Check: $12 \times 2 + 18 \times 4 = 24 + 72 = 96$. Each sentence of the first argument is one line of the second.

**Problem 4.** Solve $4x + 3y = 5$ and $6x - 5y = 17$.

**Solution.** To cancel $x$, scale so both equations carry $12x$: three times the first is $12x + 9y = 15$, and twice the second is $12x - 10y = 34$. Subtract the second of these from the first: $9y - (-10y) = 15 - 34$, so $19y = -19$ and $y = -1$.

Put this into the first original equation: $4x - 3 = 5$, so $x = 2$.

Check in the second: $6 \times 2 - 5 \times (-1) = 12 + 5 = 17$. The solution is $(2, -1)$.

**Problem 5.** For each system, compute $D = a_1b_2 - a_2b_1$ and decide how many solutions there are: (a) $2x - y = 4$, $-4x + 2y = 3$; (b) $2x - y = 4$, $-4x + 2y = -8$; (c) $2x - y = 4$, $x + 2y = 7$.

**Solution.** (a) $D = 2 \times 2 - (-4) \times (-1) = 4 - 4 = 0$. Adding twice the first equation to the second gives $0 = 11$, which is false. No solution: the lines are parallel.

(b) $D = 0$ again. Adding twice the first to the second gives $0 = 0$. The second equation is $-2$ times the first, so the lines are the same and there are infinitely many solutions, namely all points with $y = 2x - 4$.

(c) $D = 2 \times 2 - 1 \times (-1) = 5 \ne 0$, so there is exactly one solution. Twice the first plus the second gives $5x = 15$, so $x = 3$, and then $y = 2 \times 3 - 4 = 2$. Check: $3 + 2 \times 2 = 7$.

**Problem 6.** A boat travels 24 km downstream in 2 hours, and takes 3 hours to return upstream. Assume the boat's speed through the water and the speed of the current are both constant. Find them.

**Solution.** Let $b$ be the boat's speed through the water and $w$ the speed of the current, in km per hour. Going downstream the current helps, so the speed over the ground is $b + w = \frac{24}{2} = 12$. Going upstream it hinders: $b - w = \frac{24}{3} = 8$.

Adding the equations gives $2b = 20$, so $b = 10$. Subtracting gives $2w = 4$, so $w = 2$.

Check: downstream at $10 + 2 = 12$ km/h for 2 hours is 24 km, and upstream at $10 - 2 = 8$ km/h for 3 hours is 24 km. The boat makes 10 km/h and the current runs at 2 km/h.

**Problem 7.** For which value of $k$ does the system $2x + 3y = 6$, $4x + ky = 5$ have no solution? For that $k$, is there any right-hand side in place of 5 that would give infinitely many solutions?

**Solution.** The system fails to have exactly one solution when $D = 2k - 4 \times 3 = 2k - 12 = 0$, that is, when $k = 6$. For $k = 6$, subtracting twice the first equation from the second gives $0 = 5 - 12 = -7$, which is false, so there is no solution. For every other $k$ there is exactly one.

With $k = 6$ and a right-hand side $s$ in place of 5, the same subtraction gives $0 = s - 12$. This is true exactly when $s = 12$, and then the second equation, $4x + 6y = 12$, is twice the first, so the lines coincide and there are infinitely many solutions.

**Problem 8.** Find the error. "Take the system $x + y = 5$, $2x + 2y = 10$. Subtracting twice the first equation from the second gives $0 = 0$. From the first equation $y = 5 - x$, and choosing $x = 1$ gives $y = 4$. Therefore the solution of the system is $(1, 4)$."

**Solution.** Every computation is correct, and $(1, 4)$ is a solution: $1 + 4 = 5$ and $2 + 8 = 10$. The error is the word "the". The second equation is twice the first, so $D = 1 \times 2 - 2 \times 1 = 0$ and the combined equation $0 = 0$ imposes no condition. The system has the same solutions as $x + y = 5$ alone, an entire line of them, and $x = 1$ was a free choice. Choosing $x = 2$ gives $(2, 3)$, which is just as good. The correct conclusion is that the solutions are all pairs $(x, 5 - x)$.