Systems in Disguise

The choice between substitution and elimination comes down to coefficients, and a coefficient is only readable once an equation is written $a x + b y = c$ . Plenty of systems are not written that way. Every system in this lesson is an ordinary one after a cleanup move from 4.4, and the new fact is that each of the two equations is cleaned on its own.

Problem

Eliminating

y

from

6 (x - 2 y) + 7 y = 29

and

4 x + 9 y = 155

means comparing the two coefficients of

y

. In the second equation that coefficient is

9

. Enter the coefficient of

y

in the first equation, once its left side is written as

a x + b y

Show a hint

To eliminate $y$ you compare the two coefficients of $y$ , and that coefficient is whatever multiplies $y$ once the left side is collected into one $x$ -term and one $y$ -term.
Distribute the $6$ over $x - 2 y$ to get $6 x - 12 y$ , then combine that with the $+ 7 y$ outside the parentheses.

Show the full solution

Distributing gives

6 x - 12 y + 7 y

, and

- 12 y + 7 y = - 5 y

, so the left side is

6 x - 5 y

and the coefficient of

y

- 5

. The printed

7

is not the coefficient on its own, since distributing the

6

produces a second

y

-term,

- 12 y

, that combines with it.

Problem

Distributing and combining like terms turns

6 (x - 2 y) + 7 y = 29

into

6 x - 5 y = 29

, so the system is

6 x - 5 y = 29

and

4 x + 9 y = 155

. Eliminate

x

the 5.3 way, back-substitute, and enter

x

Show a hint

Nothing about elimination changes now that both equations are in $a x + b y = c$ form. Compare the two coefficients on $x$ before scaling anything.
The $x$ -coefficients $6$ and $4$ both divide $12$ , so double the first equation and triple the second.

Show the full solution

Doubling the first equation gives

12 x - 10 y = 58

and tripling the second gives

12 x + 27 y = 465

. Subtracting the first of those from the second leaves

37 y = 407

, so

y = 11

, and then

6 x - 55 = 29

gives

6 x = 84

and

x = 14

. Check,

6 (14) - 5 (11) = 84 - 55 = 29

and

4 (14) + 9 (11) = 56 + 99 = 155

. Only the cleanup was new. Once both equations are in

a x + b y = c

form, elimination is the same work as in 5.3.

Problem

Neither equation of

4 (x + 3 y) - 5 y = 123

and

6 x - 3 (y - 4) = 75

is in

a x + b y = c

form yet. Clean each one, solve the system, and enter the product

x y

. In the second equation the factor in front of the parentheses is

- 3

, and it reaches both terms inside, signs included.

Show a hint

Expand and combine on the left of each equation the 2.1 way before a solving method is picked. The constant that comes out of the second expansion belongs on the right side with the $75$ .
$- 3 (y - 4)$ is $- 3 y + 12$ , not $- 3 y - 12$ , so the second equation cleans to $6 x - 3 y = 63$ and then $2 x - y = 21$ . Isolating $y$ there costs one move.

Show the full solution

Distributing gives

4 x + 12 y - 5 y = 123

, so

4 x + 7 y = 123

, and

6 x - 3 y + 12 = 75

, so

6 x - 3 y = 63

and

2 x - y = 21

. Then

y = 2 x - 21

, and

4 x + 7 (2 x - 21) = 123

gives

18 x = 270

x = 15

y = 9

, and

x y = 135

. The

- 3

reaches the

- 4

as well, which is why the constant lands as

+ 12

Problem

Neither

9 x - 33 = 3 y + 132

nor

4 y + 5 x = 2 x + 20

reads

a x + b y = c

, since one has a number on the left and each has letter terms on both sides. Use 4.3's balance moves on each, then solve the cleaned system and enter

y

Show a hint

Take one equation at a time. Move every letter term to the left side and every number to the right side, doing the same thing to both sides at each step.
The first equation collects to $9 x - 3 y = 165$ , and dividing both sides by $3$ gives $3 x - y = 55$ . The second collects to $3 x + 4 y = 20$ , so the $x$ -columns match.

Show the full solution

Collecting the first equation gives

9 x - 3 y = 165

, and dividing both sides by

3

gives

3 x - y = 55

. The second collects to

3 x + 4 y = 20

. Subtracting the cleaned first equation from the cleaned second leaves

5 y = - 35

, so

y = - 7

, with

3 x = 55 - 7 = 48

and

x = 16

. Check

(16, - 7)

in both originals,

144 - 33 = 111 = - 21 + 132

and

- 28 + 80 = 52 = 32 + 20

. Once both equations read

a x + b y = c

, one subtraction finished the system.

Problem

Clean

7 x + 4 y = 2 x + 3 y + 86

and

4 (x + 2 y) - x - 3 y = 144

into

a x + b y = c

form. Exactly one of the two letters can then be isolated with no division at all. Enter the value of that letter.

Show a hint

Nothing can be compared until both equations are in $a x + b y = c$ form. Combine like terms in the first, and distribute the $4$ in the second.
The cleaned system is $5 x + y = 86$ and $3 x + 5 y = 144$ . Scan its four coefficients for a $1$ .

Show the full solution

Combining like terms in the first equation gives

5 x + y = 86

, and distributing in the second gives

4 x + 8 y - x - 3 y = 144

, so

3 x + 5 y = 144

. The only coefficient of

1

is on the

y

of the first equation, so

y = 86 - 5 x

, and

3 x + 5 (86 - 5 x) = 144

becomes

430 - 22 x = 144

, so

22 x = 286

and

x = 13

. Then

y = 86 - 65 = 21

. Eliminating

y

means scaling one equation to

5 y

, and eliminating

x

means scaling both to

15 x

, so

y

is the cheap letter either way.

4.4 cleared a fractional equation by multiplying both sides by a common denominator of its fractions, and one well-chosen multiplication removed every fraction at once. A system has two equations to run that on. The next two problems put fractions in one equation, then in both.

Problem

Solve

\frac{x}{3} + \frac{y}{2} = 17

and

4 x - 3 y = 60

. Only one of the two equations has fractions in it, and the other already reads

a x + b y = c

. Enter

x

Show a hint

The second equation has no fractions in it, so it needs no multiplier at all. Leave it as it is and work on the first.
The denominators $3$ and $2$ both divide $6$ . Multiply every term of the first equation by $6$ , the $17$ included.

Show the full solution

Multiplying the first equation by

6

gives

2 x + 3 y = 102

, and adding

4 x - 3 y = 60

leaves

6 x = 162

, so

x = 27

, with

3 y = 102 - 54 = 48

and

y = 16

. Check,

9 + 8 = 17

and

108 - 48 = 60

. Each equation is cleaned on its own, so only the fractional one is multiplied.

Problem

Solve

\frac{x}{10} + \frac{y}{3} = 15

and

\frac{x}{9} + \frac{y}{2} = 19

. Each equation has its own pair of denominators, so clear the fractions one equation at a time. Enter

x

Show a hint

Clearing the first equation takes a multiple of $10$ and $3$ . Clearing the second takes a multiple of $9$ and $2$ . Choose the two multipliers separately.
Multiply the first equation by $30$ and the second by $18$ , every term on both sides.

Show the full solution

Multiplying the first equation by

30

gives

3 x + 10 y = 450

, and multiplying the second by

18

gives

2 x + 9 y = 342

. Doubling the first result gives

6 x + 20 y = 900

, tripling the second gives

6 x + 27 y = 1026

, and subtracting leaves

7 y = 126

, so

y = 18

. Then

3 x + 180 = 450

gives

3 x = 270

and

x = 90

. The multiplier for each equation is built from that equation's own denominators, so the two multipliers are different.

The multiplier for a row comes from the denominators in that row and nothing else, so

24

clears the first and

35

clears the second. Each multiplier reaches every term on both sides of its own row, and neither one clears the other row.

Problem

A student clears

\frac{x}{3} + 4 = \frac{y}{9}

by multiplying by

9

and gets

3 x + 4 = y

. Paired with

5 x + y = 124

that line gives

(15, 49)

, and the check fails, since

\frac{15}{3} + 4 = 9

while

\frac{49}{9}

is not

9

. Redo the clearing and enter the correct

y

Show a hint

Multiplying by $9$ has to multiply every term, and the first equation has three terms, not two.
The $+ 4$ sits on the same side as $\frac{x}{3}$ , so it gets multiplied as well and becomes $36$ .

Show the full solution

Multiplying every term by

9

gives

3 x + 36 = y

. Substituting into

5 x + y = 124

gives

5 x + 3 x + 36 = 124

, so

8 x = 88

x = 11

, and

y = 3 (11) + 36 = 69

. Check,

\frac{11}{3} + 4 = \frac{23}{3}

and

\frac{69}{9} = \frac{23}{3}

. Every term means the

+ 4

as well, so

3 x + 4 = y

is a different equation, and the pair it gives fails the original.

A decimal coefficient is a count of tenths or hundredths, so a decimal equation is really a fraction equation, and one multiplication clears it just as before. The power of ten still comes from one equation at a time, so the two multipliers need not match.

Problem

Solve

0.6 x + 0.4 y = 37

and

0.07 x - 0.05 y = 1.9

. The power of ten that clears an equation comes from the finest decimal place in that equation. Enter

x

Show a hint

The first equation has nothing finer than tenths, and the second has hundredths, so the two powers of ten are not the same.
Multiply the first equation by $10$ and the second by $100$ , both sides each time. The right sides then read $370$ and $190$ .

Show the full solution

Clearing gives

6 x + 4 y = 370

and

7 x - 5 y = 190

. Scaling to

20 y

and

- 20 y

gives

30 x + 20 y = 1850

and

28 x - 20 y = 760

, and adding leaves

58 x = 2610

, so

x = 45

, with

y = 25

. Check,

27 + 10 = 37

and

3.15 - 1.25 = 1.9

. Multiplying the second equation by

10

would leave

0.7 x

and

0.5 y

behind.

One shape from 4.4 is worth naming again. A fraction bar is a grouping symbol, so a numerator with two terms is grouped as a whole and needs parentheses the moment it is multiplied. Inside a system that numerator carries two letters instead of one, and that is the only difference from 4.4.

Problem

Solve

\frac{4 x - y}{5} = \frac{x + 2 y}{4}

and

x + y = 125

. Each side of the first equation is a single fraction and nothing else, so two multiplications clear it. Enter

x

Show a hint

This is 4.4's shape with two letters in each numerator instead of one. The fraction bar groups its whole numerator.
Multiplying by $4$ and then by $5$ leaves $4 (4 x - y) = 5 (x + 2 y)$ . Collect that into $a x + b y = c$ before using $x + y = 125$ .

Show the full solution

Multiplying by

4

and then by

5

gives

4 (4 x - y) = 5 (x + 2 y)

, so

16 x - 4 y = 5 x + 10 y

and

11 x = 14 y

. With

x = 125 - y

11 (125 - y) = 14 y

gives

25 y = 1375

y = 55

, and

x = 70

. Check,

70 + 55 = 125

, and both sides of the first equation come out

45

, from

\frac{225}{5}

and

\frac{180}{4}

. In

a x + b y = c

form the cleaned equation is

11 x - 14 y = 0

, so

c

is allowed to be

0

Some systems are not linear in $x$ and $y$ at all. In $\frac{9}{x} + \frac{2}{y} = 41$ the letters are in the denominators, and multiplying through by $x y$ produces an $x y$ term rather than a linear equation. Both equations are linear in $\frac{1}{x}$ and $\frac{1}{y}$ though, and those two pieces can be given names.

Problem

\frac{4}{x} + \frac{3}{y} = 44

and

\frac{7}{x} - \frac{3}{y} = 11

, set

u = \frac{1}{x}

and

v = \frac{1}{y}

. Solve the system in

u

and

v

and enter

u

Show a hint

Rewrite every $\frac{1}{x}$ as $u$ and every $\frac{1}{y}$ as $v$ , so $\frac{4}{x}$ is $4 u$ and $\frac{3}{y}$ is $3 v$ .
The renamed system is $4 u + 3 v = 44$ and $7 u - 3 v = 11$ , an ordinary 5.3 problem with opposite $v$ -coefficients.

Show the full solution

u

and

v

the system is

4 u + 3 v = 44

and

7 u - 3 v = 11

. The

v

-coefficients are opposites, so adding gives

11 u = 55

and

u = 5

. Then

20 + 3 v = 44

gives

v = 8

, and since

u = \frac{1}{x}

and

v = \frac{1}{y}

, the original values are

x = \frac{1}{5}

and

y = \frac{1}{8}

Problem

Solve

\frac{14}{x} + \frac{6}{y} = 5

and

\frac{21}{x} - \frac{4}{y} = 1

. Neither equation has the form

a x + b y = c

, but naming the two repeated pieces yourself turns both into that form. Solve the renamed system and enter

x

itself, not the value of the new letter.

Show a hint

Neither equation is linear in $x$ and $y$ , but both are linear in $\frac{1}{x}$ and $\frac{1}{y}$ . Name those two pieces.
With $u = \frac{1}{x}$ and $v = \frac{1}{y}$ the system is $14 u + 6 v = 5$ and $21 u - 4 v = 1$ . Solve that by elimination, then convert with $x = \frac{1}{u}$ .

Show the full solution

With

u = \frac{1}{x}

and

v = \frac{1}{y}

the system is

14 u + 6 v = 5

and

21 u - 4 v = 1

. Doubling the first and tripling the second gives

28 u + 12 v = 10

and

63 u - 12 v = 3

, and adding leaves

91 u = 13

, so

u = \frac{1}{7}

and

v = \frac{1}{2}

. Then

x = \frac{1}{u} = 7

, with

y = 2

. Check in the originals,

\frac{14}{7} + \frac{6}{2} = 5

and

\frac{21}{7} - \frac{4}{2} = 1

Every system in this lesson was one cleanup move away from an ordinary system, and the choice between substitution and elimination came after that move rather than before it. 5.5 starts further back. The equations are not written down at all, only English sentences, and building the system is the work.

Practice these ideas

Practice

The first equation of the system

3 (x + 2 y) = 54

and

x - y = 6

is not in

a x + b y = c

form yet. Rewrite it, solve the system, and enter

x

Show the solution

Dividing the first equation by

3

gives

x + 2 y = 18

. Subtracting

x - y = 6

leaves

3 y = 12

, so

y = 4

. Back-substituting into

x - y = 6

gives

x = 10

. Check,

3 (10 + 8) = 54

and

10 - 4 = 6

. Distributing to

3 x + 6 y = 54

and then dividing by

3

lands on the same equation, so either route cleans it.

Practice

The system is

8 x - 4 (y - 3) = 52

and

2 x + y = 34

. Clean the first equation into

a x + b y = c

form, then solve the system. The ask is

y

, not

x

Show the solution

Distributing gives

8 x - 4 y + 12 = 52

, so

8 x - 4 y = 40

and

2 x - y = 10

. Adding

2 x + y = 34

leaves

4 x = 44

, so

x = 11

and

y = 12

. Check

(11, 12)

in the original first equation,

88 - 4 (9) = 52

. The product of

- 4

and

- 3

+ 12

, and dropping that sign change is the usual slip here.

Practice

Both sides of

9 x - y = 5 x + 3 y + 44

carry letters. Put that equation into

a x + b y = c

form, solve the system it makes with

2 x + 5 y = 120

, and enter

x

Show the solution

Collecting the letters on the left gives

4 x - 4 y = 44

, and dividing by

4

leaves

x - y = 11

, so

x = y + 11

. Then

2 (y + 11) + 5 y = 120

collects to

7 y = 98

, so

y = 14

and

x = 25

. Check,

9 (25) - 14 = 211

and

5 (25) + 3 (14) + 44 = 211

. In

x - y = 11

the letter

x

has coefficient

1

, so isolating it brings in no fractions.

Practice

The equation

12 + 3 x = 5 y - 8

is not in

a x + b y = c

form. Rewrite it in that form, solve it together with

2 x + 3 y = 69

, and enter the value of

y

Show the solution

Subtract

12

from both sides and

5 y

from both sides to get

3 x - 5 y = - 20

. Multiply that by

2

for

6 x - 10 y = - 40

, multiply

2 x + 3 y = 69

3

for

6 x + 9 y = 207

, and subtract the first of those from the second for

19 y = 247

, so

y = 13

. Then

3 x = 45

and

x = 15

. In the original,

12 + 45 = 57

and

65 - 8 = 57

Practice

Solve

\frac{x}{4} + \frac{y}{5} = 14

and

3 x - 2 y = 36

. Only one of the two equations has fractions to clear. Enter

x

Show the solution

Multiplying the first equation by

20

gives

5 x + 4 y = 280

. Doubling

3 x - 2 y = 36

gives

6 x - 4 y = 72

, and adding leaves

11 x = 352

, so

x = 32

. Back-substituting gives

y = 30

. Check,

\frac{32}{4} + \frac{30}{5} = 8 + 6 = 14

and

96 - 60 = 36

. Only the equation with denominators needs a multiplier, and it has to reach the right side, which is why the

14

becomes

280

Practice

Solve

\frac{x + 2 y}{3} - \frac{y}{4} = 9

and

2 x - y = 26

. Multiply the first equation by the least common denominator to clear the fractions, then solve the cleaned system. Enter

x

Show the solution

Multiplying the first equation by

12

gives

4 (x + 2 y) - 3 y = 108

, so

4 x + 5 y = 108

. The second equation gives

y = 2 x - 26

, so

4 x + 10 x - 130 = 108

and

14 x = 238

, so

x = 17

. Then

y = 8

. Check in the first equation,

\frac{17 + 16}{3} - \frac{8}{4} = 11 - 2 = 9

. Multiply the whole numerator and not just the

x

, which is why

x + 2 y

stays in parentheses until the distributing step.

Practice

Solve

\frac{x}{2} + \frac{y}{7} = 13

and

\frac{x}{8} + \frac{y}{5} = 9

. The denominators differ from one equation to the other, so clear each equation with its own multiplier. Enter

x

Show the solution

Multiplying the first equation by

14

gives

7 x + 2 y = 182

, and the second by

40

gives

5 x + 8 y = 360

. Four times the first is

28 x + 8 y = 728

, and subtracting

5 x + 8 y = 360

leaves

23 x = 368

, so

x = 16

. Then

y = 35

. Check,

\frac{16}{8} + \frac{35}{5} = 9

. The smallest multiplier for each equation comes from its own denominators.

Practice

Solve

0.8 x + 0.5 y = 18

and

0.2 x - 0.3 y = 6.2

. Clear the decimals from each equation first, then eliminate. Enter

y

Show the solution

Multiplying each equation by

10

gives

8 x + 5 y = 180

and

2 x - 3 y = 62

. Four times the second is

8 x - 12 y = 248

, and subtracting that from the first leaves

17 y = - 68

, so

y = - 4

. Back-substitution gives

x = 25

. Check,

0.8 (25) + 0.5 (- 4) = 20 - 2 = 18

and

0.2 (25) - 0.3 (- 4) = 5 + 1.2 = 6.2

Practice

Solve

0.7 x + 0.4 y = 45

and

0.03 x + 0.08 y = 3.5

. Clear the decimals in each equation first, then solve. Enter

x

Show the solution

Multiplying the first equation by

10

gives

7 x + 4 y = 450

, and the second by

100

gives

3 x + 8 y = 350

. Doubling

7 x + 4 y = 450

gives

14 x + 8 y = 900

, and subtracting

3 x + 8 y = 350

leaves

11 x = 550

, so

x = 50

. Then

y = 25

. Check in the second equation,

0.03 (50) + 0.08 (25) = 3.5

. A multiplier of

10

there would leave

0.3 x

, since a hundredths place needs a multiplier of

100

Practice

The system is

0.25 x + 0.4 y = 17

and

x - y = 3

. A student clears the first equation by multiplying by

100

and writes

25 x + 40 y = 17

, leaving the right side alone. That line has no whole-number solution with

x - y = 3

at all, which is the first sign something is wrong. Redo the clearing and enter the correct

x

Show the solution

Multiplying every term by

100

gives

25 x + 40 y = 1700

, and dividing by

5

gives

5 x + 8 y = 340

. With

x = y + 3

5 (y + 3) + 8 y = 340

gives

13 y = 325

, so

y = 25

and

x = 28

. Check,

0.25 (28) + 0.4 (25) = 7 + 10 = 17

and

28 - 25 = 3

. Multiplying only the left side changes the equation, since

25 x + 40 y = 17

is the same as

0.25 x + 0.4 y = 0.17

Practice

The system is

\frac{x}{2} - 6 = \frac{y}{8}

and

2 x + y = 138

. A student multiplies the first equation by

8

, writes

4 x - 6 = y

, and reports

x = 24

. Testing

(24, 90)

in the first equation gives

6

on the left and

11.25

on the right. Enter the correct value of

x

Show the solution

Multiplying every term by

8

gives

4 x - 48 = y

, so

4 x - y = 48

. Adding

2 x + y = 138

gives

6 x = 186

, so

x = 31

. The student multiplied only the two fraction terms and left the

- 6

unchanged. With

y = 76

, the check is

15.5 - 6 = 9.5

and

\frac{76}{8} = 9.5

Practice

Solve

\frac{2 x + y}{4} = \frac{x + 5 y}{6}

and

x + y = 99

. Clean the first equation into

a x + b y = c

before using the second. Enter

y

Show the solution

Multiplying both sides by

4

and then by

6

gives

6 (2 x + y) = 4 (x + 5 y)

, so

12 x + 6 y = 4 x + 20 y

and the cleaned form is

8 x - 14 y = 0

, or

4 x = 7 y

. With

x = 99 - y

4 (99 - y) = 7 y

gives

396 - 4 y = 7 y

, so

11 y = 396

and

y = 36

. Then

x = 63

, and the original equation checks,

\frac{162}{4} = \frac{243}{6} = 40.5

. The fraction bar groups the whole numerator, so each numerator needs parentheses once the denominators are gone.

Practice

Solve

\frac{x - y}{4} = \frac{y}{6} + 3

and

x + y = 52

. The right side of the first equation is a fraction plus a whole number rather than a single fraction, so clear it with one multiplier instead of cross-multiplying. Enter

x

Show the solution

Multiplying the first equation by

12

gives

3 (x - y) = 2 y + 36

, so

3 x - 5 y = 36

. With

x = 52 - y

3 (52 - y) - 5 y = 36

gives

156 - 8 y = 36

, so

8 y = 120

y = 15

, and

x = 37

. Check,

\frac{37 - 15}{4} = \frac{11}{2}

and

\frac{15}{6} + 3 = \frac{11}{2}

. One multiplication by

12

clears both denominators, and the

3

on the right is a term to multiply like any other.

Practice

Clean

3 (x + 2 y) - 2 y = 71

and

8 x + 9 y = 2 x + 2 y + 131

into

a x + b y = c

form. One letter's two coefficients meet at a smaller least common multiple than the other letter's, so eliminating it takes less scaling. Enter that letter.

Show the solution

Distributing gives

3 x + 6 y - 2 y = 71

, so

3 x + 4 y = 71

, and collecting the second gives

6 x + 7 y = 131

. The

x

-coefficients

3

and

6

meet at

6

, which takes one scaling, while the

y

-coefficients

4

and

7

meet at

28

, which takes two, so the cheaper letter is

x

. Both routes reach the same pair

(9, 11)

, and the only difference is how large the numbers get.

Practice

\frac{1}{x} + \frac{5}{y} = 33

and

\frac{3}{x} - \frac{1}{y} = 19

, set

u = \frac{1}{x}

and

v = \frac{1}{y}

. Solve the renamed system and enter

u

Show the solution

Renaming gives

u + 5 v = 33

and

3 u - v = 19

. Solving the second for

v

gives

v = 3 u - 19

, so

u + 5 (3 u - 19) = 33

, which is

16 u - 95 = 33

, then

16 u = 128

and

u = 8

. Then

v = 5

, and the check gives

8 + 25 = 33

and

24 - 5 = 19

. Since

u = \frac{1}{x}

and

v = \frac{1}{y}

, the original solution is

x = \frac{1}{8}

and

y = \frac{1}{5}

Practice

Solve the system

\frac{2}{x} + \frac{3}{y} = 30

and

\frac{5}{x} - \frac{2}{y} = 37

. Enter

x

as a fraction in lowest terms.

Show the solution

With

u = \frac{1}{x}

and

v = \frac{1}{y}

the system is

2 u + 3 v = 30

and

5 u - 2 v = 37

. Doubling the first gives

4 u + 6 v = 60

and tripling the second gives

15 u - 6 v = 111

, and the sum is

19 u = 171

, so

u = 9

and

v = 4

. Then

y = \frac{1}{4}

and

x = \frac{1}{u} = \frac{1}{9}

. The ask is

x

, not

u

, so

9

is one step short of the answer.

Practice

Solve

\frac{x + y}{2} + \frac{y}{3} = 23

and

0.15 x - 0.2 y = 1.5

. Clear each equation to the form

a x + b y = c

first. Enter the product

x y

Show the solution

Multiplying the first equation by

6

gives

3 (x + y) + 2 y = 138

, so

3 x + 5 y = 138

. Multiplying the second by

100

gives

15 x - 20 y = 150

, and dividing by

5

gives

3 x - 4 y = 30

. Subtracting leaves

9 y = 108

, so

y = 12

, and

3 x = 138 - 60 = 78

, so

x = 26

and

x y = 312

. Check,

\frac{38}{2} + 4 = 23

and

3.9 - 2.4 = 1.5

. Clearing fractions and clearing decimals are the same move, one multiplier taken from the denominators and one from a power of ten.

Practice

Nonnegative numbers

a

and

b

satisfy

\sqrt{a} + 3 \sqrt{b} = 41

and

\sqrt{a} - \sqrt{b} = 13

. Neither equation is linear in

a

and

b

, but both are linear in

\sqrt{a}

and

\sqrt{b}

. Name those two pieces, solve, convert back, and enter

a

Show the solution

With

u = \sqrt{a}

and

v = \sqrt{b}

the system is

u + 3 v = 41

and

u - v = 13

. Subtracting gives

4 v = 28

, so

v = 7

and

u = 20

. Then

a = u^{2} = 400

. Squaring is what converts back, so

b = 49

, and the check gives

20 + 21 = 41

and

20 - 7 = 13

. Entering

20

reports

\sqrt{a}

rather than

a

Practice

Positive numbers

m

and

n

satisfy

m^{2} + n^{2} = 289

and

m^{2} - n^{2} = 161

. The system is linear in

m^{2}

and

n^{2}

, so name those two pieces. Enter

m + n

Show the solution

With

u = m^{2}

and

v = n^{2}

the system is

u + v = 289

and

u - v = 161

. Adding gives

2 u = 450

, so

u = 225

and

v = 64

. Both numbers are positive, so

m = 15

and

n = 8

, and

m + n = 23

. The naming is what makes the system linear, since

m^{2}

and

n^{2}

appear only as whole pieces.