Substitution

Two numbers satisfy $y = 3 x + 26$ and $x + y = 2$ . Run 5.1's search on the second equation, $(0, 2)$ , $(1, 1)$ , $(2, 0)$ , and every candidate fails the first. The pair hides among negatives, where listing never ends. The first equation is not one more claim to test. It says what $y$ is. At the solution, $y$ and $3 x + 26$ are the same number.

Problem

The first equation says

y

equals

3 x + 26

, so write

3 x + 26

y

's place in

x + y = 2

. One equation in one variable remains. Solve it and enter

x

Show a hint

The swap leaves an equation with only $x$ in it, exactly the kind chapter 4 solves.
After the substitution the claim reads $x + (3 x + 26) = 2$ . Collect the left side to $4 x + 26 = 2$ .

Show the full solution

x + (3 x + 26) = 2

collects to

4 x + 26 = 2

, so

4 x = - 24

and

x = - 6

. The first equation names

y

, so its expression can stand wherever

y

stands.

Problem

The partner is still missing. Substitute

x = - 6

into

y = 3 x + 26

, the equation already solved for

y

, and enter

y

. Check the pair in both original equations first.

Show a hint

The solved-form equation computes $y$ from any $x$ , including this one.
Substituting gives $y = 3 (- 6) + 26$ . Keep the parentheses around the negative value.

Show the full solution

y = 3 (- 6) + 26 = - 18 + 26 = 8

. Check

(- 6, 8)

in both originals,

8 = 3 (- 6) + 26

and

- 6 + 8 = 2

, so the pair passes the full 5.1 test.

Problem

The system

x = 3 y + 22

and

x + 2 y = 112

leads with

x

this time. Replace

x

in the second equation with

3 y + 22

, solve for

y

, back-substitute, and enter

x

Show a hint

Either letter can be the pre-solved one. The steps do not change. Replace the letter that equals an expression.
$(3 y + 22) + 2 y = 112$ collects to $5 y + 22 = 112$ .

Show the full solution

(3 y + 22) + 2 y = 112

gives

5 y + 22 = 112

, so

5 y = 90

y = 18

, and

x = 3 (18) + 22 = 76

. The ask is

x

, so the back-substitution is not optional.

So far the replaced letter stood alone. In $4 x + 3 y = 47$ the $y$ is tripled, and $3 y$ means three copies of all of $y$ , so whatever stands in for $y$ must arrive as one unit. With $y = 2 x + 7$ , the incoming expression takes the seat as a block, not a piece at a time.

The equation

y = 2 x + 7

puts the expression

2 x + 7

on a tile, and lifting

y

out of

4 x + 3 y = 47

leaves a slot after the

3

. The tile lands as one unit, so parentheses close around it,

4 x + 3 (2 x + 7) = 47

. Substituting a piece, or dropping the parentheses, writes a different equation.

Problem

Solve the system

y = 3 x - 16

and

5 x + 2 y = 78

. The

2 y

doubles the whole expression, so substitute in parentheses, distribute, and enter

x + y

Show a hint

Substitute the whole expression $3 x - 16$ for $y$ in the second equation, with parentheses around all of it.
$5 x + 2 (3 x - 16) = 78$ distributes to $11 x - 32 = 78$ .

Show the full solution

5 x + 2 (3 x - 16) = 78

gives

11 x - 32 = 78

, so

11 x = 110

and

x = 10

, then

y = 3 (10) - 16 = 14

and

x + y = 24

Problem

A student substitutes

y = 2 x - 3

into

x + 4 y = 51

and writes

x + 4 \cdot 2 x - 3 = 51

, no parentheses. That line gives

x = 6

, then

y = 9

, and the check fails, since

6 + 4 (9) = 42

, not

51

. Redo the substitution with the parentheses it needed and enter the correct

x

Show a hint

In $4 y$ , the $4$ must multiply everything $y$ is, not just the first piece.
The faithful line is $x + 4 (2 x - 3) = 51$ . Distribute, then collect.

Show the full solution

x + 4 (2 x - 3) = 51

gives

9 x - 12 = 51

, so

9 x = 63

and

x = 7

, with

y = 2 (7) - 3 = 11

and

7 + 44 = 51

. The bare line subtracted

3

where the equation subtracts

12

Nobody pre-solves equations for you outside a lesson. When neither equation arrives in solved form, pick a letter, solve for it with 4.3's moves, then substitute exactly as before. One isolate is the whole cost, and the rest of the run does not change.

Problem

Neither equation of

3 x + y = 32

and

7 x + 2 y = 75

is solved for a letter, but the first is one 4.3 move away. Solve it for

y

, substitute into the second, and enter

y

. Check the pair in both originals.

Show a hint

Subtracting $3 x$ from both sides pre-solves the first equation for $y$ .
Substitute $y = 32 - 3 x$ into the second equation to get $7 x + 2 (32 - 3 x) = 75$ .

Show the full solution

y = 32 - 3 x

, so

7 x + 2 (32 - 3 x) = 75

collects to

x + 64 = 75

and

x = 11

, then

y = 32 - 33 = - 1

. Check,

33 + (- 1) = 32

and

77 - 2 = 75

, both originals pass.

Problem

3 x - y = 20

and

5 x + 2 y = 4

the

y

has coefficient

- 1

. Solving the first for

y

still takes one move,

y = 3 x - 20

. Substitute into the second equation and enter

y

Show a hint

Coefficient $- 1$ is as cheap as coefficient $1$ . No dividing happens on the way to solved form.
$5 x + 2 (3 x - 20) = 4$ collects to $11 x - 40 = 4$ .

Show the full solution

5 x + 2 (3 x - 20) = 4

gives

11 x - 40 = 4

, so

11 x = 44

x = 4

, and

y = 3 (4) - 20 = - 8

. Nothing stops a solution pair from being negative.

Problem

One letter in

4 x + y = 198

and

2 x + 3 y = 204

isolates without fractions. Find it, isolate it, substitute, solve the system, and enter the product

x y

Show a hint

Only $y$ in the first equation has coefficient $1$ , so isolating it costs one 4.3 move and no fractions.
Substituting $y = 198 - 4 x$ into the second equation gives $2 x + 3 (198 - 4 x) = 204$ .

Show the full solution

Isolate

y = 198 - 4 x

, then

2 x + 3 (198 - 4 x) = 204

gives

2 x + 594 - 12 x = 204

, so

- 10 x = - 390

and

x = 39

, then

y = 198 - 156 = 42

. The product is

39 \cdot 42 = 1638

. Check,

2 (39) + 3 (42) = 78 + 126 = 204

Problem

Solve the system

y = 2 x + 4

and

6 x + 2 y = 33

. Nothing requires the pair to be whole. Check the pair in both equations, then enter

x

as a fraction in lowest terms.

Show a hint

The run is the usual one. Only the ending is not a whole number.
$6 x + 2 (2 x + 4) = 33$ collects to $10 x + 8 = 33$ .

Show the full solution

6 x + 2 (2 x + 4) = 33

gives

10 x + 8 = 33

, so

10 x = 25

and

x = \frac{25}{10} = \frac{5}{2}

, with

y = 2 \cdot \frac{5}{2} + 4 = 9

. Check,

9 = 5 + 4

and

15 + 18 = 33

, both originals pass.

Sometimes the substitution erases the letter entirely. Every $x$ -term cancels, and a statement with no variable in it is left behind, true or false on its own. 4.3 met this inside one equation. Here the leftover statement is a verdict about the whole system, and the next two problems read one of each kind.

Problem

Substitute

y = 4 x + 5

into

8 x - 2 y = 6

and simplify. Every

x

-term cancels and a number statement is left. Decide what it means and enter the number of solutions of the system.

Show a hint

4.3 met a vanishing variable inside one equation. The number statement left behind is the verdict.
$8 x - 2 (4 x + 5)$ simplifies to $- 10$ , and the equation claims that equals $6$ .

Show the full solution

8 x - 2 (4 x + 5) = 8 x - 8 x - 10 = - 10

, so the substituted equation reads

- 10 = 6

, false at every

x

, and the system has

0

solutions. No pair can make both claims true at once.

Problem

Substituting

y = 2 x + 6

into

4 x - 2 y = - 12

leaves

- 12 = - 12

, true for every

x

. Infinitely many pairs solve the system, but not every pair does. Of

(2, 10)

(6, 18)

(10, 26)

(8, 20)

, enter how many solve the system.

Show a hint

Every pair solving the first equation solves both. Pairs off it solve neither.
Test each pair in $y = 2 x + 6$ , first value in for $x$ , second in for $y$ .

Show the full solution

10 = 4 + 6

18 = 12 + 6

, and

26 = 20 + 6

pass, but

2 (8) + 6 = 22 \neq 20

, so

3

of the four solve the system. Infinitely many means every pair satisfying the equation, and only those.

Problem

Two numbers satisfy

x - 2 y = 45

and

2 x + 3 y = 398

. Isolate the coefficient-1 letter, substitute, solve the system, check the pair in both equations, and enter the product

x y

Show a hint

Only $x$ carries coefficient $1$ . One 4.3 move gives $x = 2 y + 45$ .
Substituting gives $2 (2 y + 45) + 3 y = 398$ , which collects to $7 y + 90 = 398$ .

Show the full solution

Isolate

x = 2 y + 45

, then

2 (2 y + 45) + 3 y = 398

gives

7 y + 90 = 398

, so

7 y = 308

and

y = 44

, then

x = 2 (44) + 45 = 133

. Check,

133 - 88 = 45

and

266 + 132 = 398

. The product is

133 \cdot 44 = 5852

All the systems in this lesson had a letter with coefficient $1$ or $- 1$ , or were already solved for a letter. There is no letter to isolate cleanly in $9 x + 7 y = 76$ and $5 x + 8 y = 72$ , so any isolation starts with a division and brings fractions into every later step. Lesson 5.3 introduces elimination, a second method for exactly this case.

Practice these ideas

Practice

The numbers

x

and

y

satisfy

y = x + 25

and

x + y = 71

. Substitute the expression for

y

into the second equation, solve, and enter

x

Show the solution

Substituting gives

x + (x + 25) = 71

, so

2 x + 25 = 71

2 x = 46

, and

x = 23

. Then

y = 48

, and

23 + 48 = 71

checks.

Practice

The system

y = 2 x

and

3 x + y = 40

. The expression for

y

has no constant term, and the steps are exactly the same. Substitute, solve, back-substitute, and enter

y

Show the solution

Substituting gives

3 x + 2 x = 40

, so

5 x = 40

and

x = 8

. Back-substituting,

y = 2 (8) = 16

. The pair checks in the second equation, since

24 + 16 = 40

Practice

The system

x = y + 10

and

x + 4 y = 30

leads with

x

. Replace the

x

of the second equation with

y + 10

, solve for

y

, back-substitute, and enter

x

Show the solution

(y + 10) + 4 y = 30

collects to

5 y + 10 = 30

, so

5 y = 20

and

y = 4

. Then

x = 4 + 10 = 14

. Check,

14 + 4 (4) = 30

Practice

Solve the system

y = 2 x + 34

and

2 x + y = 150

. Substitute, solve for

x

, then finish with a back-substitution and enter

y

Show the solution

2 x + (2 x + 34) = 150

gives

4 x = 116

, so

x = 29

, and

y = 2 (29) + 34 = 92

. Check,

58 + 92 = 150

Practice

y = x - 5

and

3 x + 2 y = 50

, the

y

slot carries a coefficient, so the substitution needs parentheses. Substitute, distribute, solve the system, and enter

x + y

Show the solution

3 x + 2 (x - 5) = 50

distributes to

3 x + 2 x - 10 = 50

, so

5 x = 60

and

x = 12

. Then

y = 12 - 5 = 7

and

x + y = 19

. Check,

3 (12) + 2 (7) = 36 + 14 = 50

. Dropping the parentheses would give

3 x + 2 x - 5 = 50

, which subtracts

5

where the equation subtracts

10

Practice

Solve the system

y = 2 x + 1

and

5 x + 3 y = 80

. The

3 y

takes the whole expression as one unit. Substitute with parentheses, distribute, and enter

y

Show the solution

5 x + 3 (2 x + 1) = 80

distributes to

5 x + 6 x + 3 = 80

, so

11 x = 77

and

x = 7

. Then

y = 2 (7) + 1 = 15

, and

35 + 45 = 80

checks.

Practice

Substituting

y = 3 x - 7

into

2 x + 5 y = 61

, a student writes

2 x + 5 \cdot 3 x - 7

, no parentheses. At every value of

x

, the student's expression exceeds the faithful

2 x + 5 (3 x - 7)

by the same fixed amount. Enter that amount.

Show the solution

The faithful expression is

2 x + 15 x - 35 = 17 x - 35

and the bare one is

2 x + 15 x - 7 = 17 x - 7

, a gap of

35 - 7 = 28

at every

x

. The missing parentheses shrink the subtraction from

35

7

Practice

Neither equation of

4 x + y = 231

and

2 x + 3 y = 233

is pre-solved. Isolate the coefficient-1 letter with one 4.3 move, substitute into the other equation, and enter

x

Show the solution

Isolate

y = 231 - 4 x

, then

2 x + 3 (231 - 4 x) = 233

gives

2 x + 693 - 12 x = 233

, so

- 10 x = - 460

and

x = 46

. Then

y = 231 - 184 = 47

, and

92 + 141 = 233

checks.

Practice

For

x + 5 y = 14

and

3 x + 4 y = 64

, the letter that isolates cleanly is

x

. Isolate it, substitute into the second equation, solve, and enter

y

Show the solution

x = 14 - 5 y

, so

3 (14 - 5 y) + 4 y = 64

gives

42 - 15 y + 4 y = 64

, then

42 - 11 y = 64

- 11 y = 22

, and

y = - 2

, with

x = 14 + 10 = 24

. Check,

24 - 10 = 14

and

72 - 8 = 64

Practice

Solve

x = 2 y + 3

and

3 x + 2 y = 11

. The system leads with

x

, and the answer is not a whole number. Enter

y

as a fraction in lowest terms.

Show the solution

3 (2 y + 3) + 2 y = 11

gives

6 y + 9 + 2 y = 11

, so

8 y = 2

and

y = \frac{2}{8} = \frac{1}{4}

, with

x = 2 \cdot \frac{1}{4} + 3 = \frac{7}{2}

. Check,

3 \cdot \frac{7}{2} + \frac{1}{2} = 11

Practice

4 x + y = 248

and

2 x + 3 y = 254

, one letter isolates without fractions and one does not. Pick well, solve the system, and enter the product

x y

Show the solution

Isolate

y = 248 - 4 x

, then

2 x + 3 (248 - 4 x) = 254

gives

2 x + 744 - 12 x = 254

, so

- 10 x = - 490

and

x = 49

, then

y = 248 - 196 = 52

. The product is

49 \cdot 52 = 2548

. Check,

98 + 156 = 254

Practice

Substitute

y = 5 x - 3

into

10 x - 2 y = 12

and simplify. The letter vanishes, and the left side sits the same fixed amount below

12

at every

x

. Enter that amount.

Show the solution

10 x - 2 (5 x - 3) = 10 x - 10 x + 6 = 6

, so the substituted equation reads

6 = 12

, and the left side sits

12 - 6 = 6

below the right side at every

x

. The statement is false everywhere, so the system has no solution.

Practice

Substituting

y = 4 x + 1

into

12 x - 3 y = - 3

leaves

- 3 = - 3

, a true statement. Every pair solving the first equation solves the system. Of the pairs

(3, 13)

(0, 1)

(2, 8)

(5, 20)

, enter how many solve it.

Show the solution

Testing each pair in

y = 4 x + 1

13 = 12 + 1

and

1 = 0 + 1

pass, while

4 (2) + 1 = 9 \neq 8

and

4 (5) + 1 = 21 \neq 20

fail. So

2

of the four solve the system.

Practice

For

x - 2 y = 1

and

6 x + 5 y = 74

, isolate

x

, substitute into the second equation, solve the system, and enter

x + y

Show the solution

x = 2 y + 1

, so

6 (2 y + 1) + 5 y = 74

gives

12 y + 6 + 5 y = 74

, then

17 y = 68

and

y = 4

, with

x = 2 (4) + 1 = 9

. So

x + y = 13

. Check,

9 - 8 = 1

and

54 + 20 = 74

Practice

One number is 5 less than double another, and the two add to 43. In symbols,

y = 2 x - 5

and

x + y = 43

. Solve the system and enter the larger number.

Show the solution

x + (2 x - 5) = 43

gives

3 x - 5 = 43

, so

3 x = 48

and

x = 16

, then

y = 2 (16) - 5 = 27

. The larger number is

27

. The problem asks for the larger of the two numbers, not for

x

, so find both values before answering.

Practice

For

6 x + y = 37

and

4 x + 3 y = 90

, isolate the coefficient-1 letter, substitute, and enter

x

as a fraction in lowest terms.

Show the solution

y = 37 - 6 x

, so

4 x + 3 (37 - 6 x) = 90

gives

4 x + 111 - 18 x = 90

, then

111 - 14 x = 90

14 x = 21

, and

x = \frac{21}{14} = \frac{3}{2}

, with

y = 37 - 9 = 28

. The clean isolate keeps fractions out of the middle of the work, not out of the answer.

Practice

3 x - y = 106

and

2 x + 2 y = 220

, the

y

has coefficient

- 1

, and 4.3's moves still isolate it,

y = 3 x - 106

. Substitute, solve the system, and enter the product

x y

Show the solution

Substitute

y = 3 x - 106

, so

2 x + 2 (3 x - 106) = 220

gives

8 x - 212 = 220

, then

8 x = 432

and

x = 54

, with

y = 3 (54) - 106 = 56

. The product is

54 \cdot 56 = 3024

. Check,

108 + 112 = 220