unit 5 powerpoint[1] algebra (1)

66
can be written as: 6 × x × x + 25 × x × y (−5) 2 yx + 2x 2 + 4x 2 3 2 6 25 x xy x you may have noticed that x(6x + 25y) also works. 2 6 25 x xy

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Page 1: Unit 5 powerpoint[1] algebra (1)

can be written as:

6 × x × x + 25 × x × y

(−5)2yx + 2x2 + 4x2

3 26 25x x y

x

And you may have noticed that x(6x + 25y) also works.

26 25x xy

Page 2: Unit 5 powerpoint[1] algebra (1)

Evaluate the following expressions when s = 6 and t = −1.

(a) 2(s + t)

(b) s + t2

(c) st + 3t

(d) t(s + 3)

Page 3: Unit 5 powerpoint[1] algebra (1)

Evaluate the following expressions when s = 6 and t = −1.

(a) 2(s + t) = 2 × (6 + (−1)) = 2 × 5 = 10

(b) s + t2 = 6 + (−1)2 = 6 + 1 = 7

(c) st + 3t = 6 × (−1) + 3 × (−1) = −9

(d) t(s + 3) = (−1) × (6 + 3) = −9

Page 4: Unit 5 powerpoint[1] algebra (1)

Identify the terms in the expression

34 3 27k h k

.

Page 5: Unit 5 powerpoint[1] algebra (1)

Identify the terms in the expression

34 3 27k h k

The terms are +4

3k3h

27 k

Page 6: Unit 5 powerpoint[1] algebra (1)

In the expression 24 3 2a a b

(a) The coefficient of b is 2.

(b) a2 has no coefficient.

(c) The coefficient of a2 is 1.

(d) The coefficient of a2 is −1.

(e) 3 is a constant term.

(f) 3 − 2b is a term.

(g) a2 is a term.

(h) 4a and − a2 are like terms.

True or false?

Page 7: Unit 5 powerpoint[1] algebra (1)
Page 8: Unit 5 powerpoint[1] algebra (1)

Simplify the following expressions by collecting like terms.

(a) k + 3k

(b) 2m + m2 − m

(c) 6xy − 3x − 2x + y

(d) 2ab − a2b − 5ba

(e) 4a + 2A + 3 − 1

(f) 4pq − 3q + pq +3q

Page 9: Unit 5 powerpoint[1] algebra (1)

Simplify the following expressions by collecting like terms.

(a) k + 3k = 4k

(b) 2m + m2 − m = m + m2

(c) 6xy − 3x − 2x + y = 6xy − 5x + y

(d) 2ab − a2b − 5ba = −3ab − a2b

(e) 4a + 2A + 3 − 1 = 4a + 2A + 2

(f) 4pq − 3q + pq + 3q = 5pq

Page 10: Unit 5 powerpoint[1] algebra (1)

Which of the following expressions are equivalent?

(a) xy + 3x − y

(b) xy − y + 3x

(c) y + xy + 3x

(d) y − xy + 3x

(e) 3x + yx − y

Page 11: Unit 5 powerpoint[1] algebra (1)

Write the following terms in their simplest forms.

(a) a × 2 × 3 × b × a

(b) 2a × 4ab

(c) t10 × 3t5

(d) a × (−b) × (−b)

(e) (−2) × g × 3h × (−g) × 7

Page 12: Unit 5 powerpoint[1] algebra (1)

Write the following terms in their simplest forms.

(a) a × 2 × 3 × b × a = 6a2b

(b) 2a × 4ab = 8a2b

(c) t10 × 3t5 = 3t15

(d) a × (−b) × (−b) = ab2

(e) (−2) × g × 3h × (−g) × 7 = 42g2h

Page 13: Unit 5 powerpoint[1] algebra (1)

Identify the terms in the following expressions and then simplify.

(a) 5a × 2b − (−a) × 4 × b

(b) 2 × (−7P) × P + Q × 8 × P × (−2)

Page 14: Unit 5 powerpoint[1] algebra (1)

Identify the terms in the following expressions and then simplify.

(a) The terms are + 5a × 2b and − (−a) × 4 × b.

5a × 2b − (−a) × 4 × b = 10ab + 4ab = 14ab

(b) The terms are + 2 × (−7P) × P and + Q × 8 × P × (−2).

2 × (−7P) × P + Q × 8 × P × (−2) = −14P2 − 16PQ

Page 15: Unit 5 powerpoint[1] algebra (1)
Page 16: Unit 5 powerpoint[1] algebra (1)

Multiply out the brackets in the following expressions.

(a) 2a(3a − 6)

(b) 2(x + x2)y

(c) −2(4 − 2a)

(d) a(b + c) + b(a − c)

(e) 2a − (b − a)

Page 17: Unit 5 powerpoint[1] algebra (1)

Multiply out the brackets in the following expressions.

(a) 2a(3a − 6) = 6a2 − 12a

(b) 2(x + x2)y = 2xy + 2x2y

(c) −2(4 − 2a) = −8 + 4a (or 4a − 8)

(d) a(b + c) + b(a − c) = ab + ac + ab − bc = 2ab + ac − bc

(e) 2a − (b − a) = 2a − b + a = 3a − b

Page 18: Unit 5 powerpoint[1] algebra (1)

True or false?

(a) k − 3k = 2k

(b) a × b2 × (−2a) = −2a2b2

(c) 4 − (2 − x) = 2 + x

(d) a(b + c) − b(a − c) − c(b + a) = 0

Page 19: Unit 5 powerpoint[1] algebra (1)

2 6

2

A

Expand the following algebraic fractions and then simplify where possible.

(a)

(b)25 3p p q pq

p

Page 20: Unit 5 powerpoint[1] algebra (1)

2 63

2 2

2 6

2

A AA

Expand the following algebraic fractions and then simplify where possible.

(a)

(b)2 25 3

5 35 3 p pp p q pq

pq qp p p

q pq

p

Page 21: Unit 5 powerpoint[1] algebra (1)

Start with a number

Start with n

Think of a number

Double it

Add 1

Multiply by 5

Subtract 3

Subtract ten times the number you first thought of

And the answer is…

Page 22: Unit 5 powerpoint[1] algebra (1)

Start with a number

Start with n

Think of a number 5 n

Double it 10 2n

Add 1 11 2n + 1

Multiply by 5 55 5(2n + 1) = 10n + 5

Subtract 3 52 10n + 2

Subtract ten times the number you first thought of

2 (10n + 2) − 10n = 2

And the answer is… 2 2

Page 23: Unit 5 powerpoint[1] algebra (1)

Start with a number

Start with n

Think of a number

Double it

Subtract 3

Multiply by 2

Add 10

Divide by 4

Subtract 1

And the answer is…

Page 24: Unit 5 powerpoint[1] algebra (1)

Start with a number

Start with n

Think of a number −1 n

Double it −2 2n

Subtract 3 −5 2n − 3

Multiply by 2 −10 2(2n −3) = 4n − 6

Add 10 0 4n + 4

Divide by 4 0

Subtract 1 −1 n

And the answer is… −1 n

4 4 4 41

4 4 4

n nn

Page 25: Unit 5 powerpoint[1] algebra (1)

Prove that the product of an even number and a multiple of 3 is always divisible by 6. (Hint: Write the even number as 2a and the multiple of 3 as 3b, where a and b are integers.)

Page 26: Unit 5 powerpoint[1] algebra (1)

Prove that the product of an even number and a multiple of 3 is always divisible by 6. (Hint: Write the even number as 2a and the multiple of 3 as 3b, where a and b are integers.)

Let the even number be 2a and the multiple of 3 be 3b where a and b are integers.

Their product is (2a) × (3b) = 6 × ab.

ab is the product of integers and so is also an integer.

6 × ab is a multiple of 6 and so is divisible by 6.

Page 27: Unit 5 powerpoint[1] algebra (1)

Do you think that the sum of an even number and a multiple of 3 is divisible by 6?

Find a counterexample, that is, find two numbers, an even number and a multiple of 3, whose sum is not divisible by 6.

Page 28: Unit 5 powerpoint[1] algebra (1)

Do you think that the sum of an even number and a multiple of 3 is divisible by 6?

Find a counterexample, that is, find two numbers, an even number and a multiple of 3, whose sum is not divisible by 6.

A counterexample could be 4 (an even number) and 9 (a multiple of 3). Their sum, 13, is not divisible by 6.

Page 29: Unit 5 powerpoint[1] algebra (1)
Page 30: Unit 5 powerpoint[1] algebra (1)

Is x = 5 a solution to these equations?

(a) x + 1 = 4

(b) 3x − 4 = 1 + 2x

(c) 2(x − 1) − (13 − x) = 0

Page 31: Unit 5 powerpoint[1] algebra (1)

Is x = 5 a solution to these equations?

(a) x + 1 = 4

When x = 5, LHS = 5 + 1 = 6 ≠ 4 So x = 5 is not a solution to the equation.

(b) 3x − 4 = 1 + 2x

When x = 5, LHS = 3 × 5 − 4 = 11 and RHS = 1 + 2 × 5 = 11 = LHS. So x = 5 is a solution to the equation.

(c) 2(x − 1) − (13 − x) = 0

When x = 5, LHS = 2 × (5 − 1) − (13 − 5) = 2 × 4 − 8 = 0 = RHS. So x = 5 is a solution to the equation.

Page 32: Unit 5 powerpoint[1] algebra (1)

Show (without solving the equation) that x = −2 is a solution to the equation 3 − 5x = 15 + x. Now solve the equation.

Page 33: Unit 5 powerpoint[1] algebra (1)

Show (without solving the equation) that x = −2 is a solution to the equation 3 − 5x = 15 + x. Now solve the equation.

When x = −2, LHS = 3 − 5 × (−2) = 13 and RHS = 15 + (−2) = 13. LHS = RHS so −2 is a solution to the equation.

The equation is: 3 − 5x = 15 + xAdd 5x: 3 = 15 + 6xSwap sides: 15 + 6x = 3Subtract 15: 6x = −12Divide by 2: x = −2The solution is x = −2

Page 34: Unit 5 powerpoint[1] algebra (1)

Solve the following equations.

(a) 2a = 22

(b) a + 3 = 8

(c)

(d) 2k = 7 + k

(e) y = 12 − 2y

13

x

Page 35: Unit 5 powerpoint[1] algebra (1)

(a) The equation is: 2a = 22Divide by 2: a = 11The solution is a = 11.

(Check: if a = 11, LHS = 2 × 11 = 22 = RHS; the solution is correct.)

(b) The equation is: a + 3 = 8Subtract 3: a = 5The solution is a = 5.

(Check: if x = 5, LHS = 5 + 3 = 8 = RHS; the solution is correct.)

(c) The equation is:

Multiply by 3: x = 3The solution is x = 3.

(Check: if x = 3, LHS = 3/3 = 1 = RHS; the solution is correct.)

13

x

Page 36: Unit 5 powerpoint[1] algebra (1)

(d) The equation is: 2k = 7 + kSubtract k: k = 7The solution is k = 7.

(Check: if k = 7, LHS = 2 × 7 = 14 and RHS = 7 + 7 = 14, so LHS = RHS and the solution is correct.)

(e) The equation is: y = 12 − 2yAdd 2y: 3y = 12Divide by 3: y = 4The solution is y = 4.

(Check: if y = 4, LHS = 4 and RHS = 12 − 2 × 4 = 4, so LHS = RHS and the solution is correct.)

Page 37: Unit 5 powerpoint[1] algebra (1)

Solve the following equations.

(a) B + 13 = 7 − 5B

(b) x − 1 = 2(x + 1)

(c)

(d)

(e)

31

6

x

3( 7) 142

yy

51

3 6

a a

Page 38: Unit 5 powerpoint[1] algebra (1)

(a) The equation is: B + 13 = 7 − 5BAdd 5B: 6B + 13 = 7Subtract 13: 6B = −6Divide by 6: B = −1The solution is B = −1.

(Check: if B = −1, LHS = −1 + 13 = 12 and RHS = 7 − 5 × (−1) = 12, so LHS = RHS and the solution is correct.)

(b) The equation is: x − 1 = 2(x + 1)Multiply out the bracket: x − 1 = 2x + 2Subtract x: −1 = x + 2Swap sides: x + 2 = −1Subtract 2: x = −3The solution is x = −3.

(Check: if x = −3, LHS = −3 − 1 = −4 and RHS = 2 × (−3 + 1) = −4, so LHS = RHS and the solution is correct.)

Page 39: Unit 5 powerpoint[1] algebra (1)

(c) The equation is:

Multiply by 6: x + 3 = 6Subtract 3: x = 3The solution is x = 3.

(Check: if x = 3, LHS = = 1 = RHS; the solution is correct.)

(d) The equation is:

Multiply by 2: 6(y − 7) = 28 − yMultiply out the bracket: 6y − 42 = 28 − yAdd y: 7y − 42 = 28Add 42: 7y = 70Divide by 10: y = 10The solution is y = 10.

(Check: if y = 10, LHS = 3 × (10 − 7) = 9 and RHS = 14 − = 9, so LHS = RHS and the solution is correct.)

31

6

x

3( 7) 142

yy

10

2

3 3

6

Page 40: Unit 5 powerpoint[1] algebra (1)

(e) The equation is:

Multiply by 6: 2a = a − 5 + 6Tidy up: 2a = a + 1Subtract a: a = 1The solution is a = 1.

(Check: if x = 1, LHS = and RHS = ,

so LHS = RHS and the solution is correct.)

51

3 6

a a

1

3

1 5 4 11 1

6 6 3

Page 41: Unit 5 powerpoint[1] algebra (1)
Page 42: Unit 5 powerpoint[1] algebra (1)

A table has been reduced in price by 25%. It now costs £270. Express this information as an equation and then solve the equation to find how much the table cost before the reduction.How would your equation need to change if the price reduction was 20%?

Page 43: Unit 5 powerpoint[1] algebra (1)

A table has been reduced in price by 25%. It now costs £270. Express this information as an equation and then solve the equation to find how much the table cost before the reduction.How would your equation need to change if the price reduction was 20%?

Let the original price of the table be £P.The price has been reduced by 25%, so the new price is 75% of £P, or £0.75P.The table now costs £270. So 0.75P = 270.

Divide by 0.75: 270360

0.75P

The cost of the table before the reduction was £360.

(Check: 75% of £360 = 0.75 × £360 = £270.)

If the reduction was 20%, the equation would become 0.8P = 270.

Page 44: Unit 5 powerpoint[1] algebra (1)

Sara is five times the age she was 28 years ago. How old is Sara?

Page 45: Unit 5 powerpoint[1] algebra (1)

Sara is five times the age she was 28 years ago. How old is Sara?

Let Sara’s age now be A.We’re told that Sara’s age 28 years ago × 5 = her age now.28 years ago, Sara’s age was A − 28.So the equation is 5 × (A − 28) = AMultiply out the bracket: 5A − 140 = ASubtract A: 4A − 140 = 0Add 140: 4A = 140Divide by 4: A = 35So Sara is 35 years old.

(Check: 28 years ago, Sara was 7 years old and 5 times 7 gives 35, her present age.)

Page 46: Unit 5 powerpoint[1] algebra (1)

In four years time, Tom will be twice the age he was 11 years ago. How old is Tom?

Page 47: Unit 5 powerpoint[1] algebra (1)

In four years time, Tom will be twice the age he was 11 years ago. How old is Tom?

Let Tom’s age now be T.We’re told that in four years time, Tom will be twice the age he was 11 years ago.

Putting this in a table: Time Tom’s age

Now T

In 4 years time T + 4

11 years ago T − 11

Create an equation: T + 4 = 2 × (T − 11)Multiply out the bracket: T + 4 = 2T − 22Subtract T: 4 = T − 22Swap sides: T − 22 = 4Add 22: T = 26So Tom is 26 years old now.

(Check: in 4 years time Tom will be 30. 11 years ago, he was 15, and twice 15 gives 30.)

Page 48: Unit 5 powerpoint[1] algebra (1)

A man is five times as old as his son. In two years time, he will only be four times as old as his son. How old is the son now?

Page 49: Unit 5 powerpoint[1] algebra (1)

A man is five times as old as his son. In two years time, he will only be four times as old as his son. How old is the son now?

Let the son’s age now be s. The father is five times as old as his son, so he is 5s.We’re told that in two years time, he will only be four times as old as his son.

Putting this in a table:

Time Son’s age Father’s age

Now s 5s

In 2 years time

s + 2 5s + 2

Create an equation: 5s + 2 = 4 × (s + 2)Multiply out the bracket: 5s + 2 = 4s + 8Subtract 4s: s + 2 = 8Subtract 2: s = 6So the son is 6 years old now.

(Check: if the son is 6, then his father is five times as old, 30. In two years time, the son will be 8 and the father will be 32, which is 4 times the son’s age.)

Page 50: Unit 5 powerpoint[1] algebra (1)

Bob has 2 children, one two years older than the other. At present Bob is twice as old as the sum of his children’s ages. In 14 years time, Bob’s age will be exactly the sum of the ages of his two children. How old are the children now?

Page 51: Unit 5 powerpoint[1] algebra (1)

Bob has 2 children, one two years older than the other. At present Bob is twice as old as the sum of his children’s ages. In 14 years time, Bob’s age will be exactly the sum of the ages of his two children. How old are the children now?

Let the younger child’s age be x. Then the older child’s age is x + 2. Bob is twice as old as the sum of their ages, so Bob’s age is 2 × (x + x + 2) = 2(2x + 2).We’re told that in 14 years time, Bob’s age will be exactly the sum of the ages of his two children. Putting this in a table:

Time Younger child’s age Older child’s age Bob’s age

Now x x + 2 2(2x + 2)

In 14 years time x + 14 x + 16 2(2x + 2) + 14

Create an equation: 2(2x + 2) + 14 = (x + 14) + (x + 16)Remove brackets: 4x + 4 + 14 = 2x + 30Tidy up: 4x + 18 = 2x + 30Subtract 2x: 2x + 18 = 30Subtract 18: 2x = 12Divide by 2: x = 6So the children are 6 and 8 years old.

(Check: if the children are 6 and 8, then Bob is now twice, the sum of their ages, which is 28. In 14 years time, the children will be 20 and 22 and Bob will be 42, the sum of their ages.)

Page 52: Unit 5 powerpoint[1] algebra (1)

A number is divided by 7 and then 6 is added to the result giving 15. What number did you start with?

Page 53: Unit 5 powerpoint[1] algebra (1)

A number is divided by 7 and then 6 is added to the result giving 15. What number did you start with?

Let the number you start with be n.Create an equation:

6 157

n

Multiply by 7: n + 42 = 105Subtract 42: n = 63You start with the number 63.

(Check: if the number is 63, divide it by 7 to get 9, the add 6 to get 15.)

Page 54: Unit 5 powerpoint[1] algebra (1)

The sum of three consecutive integers is 351. Find the numbers.

Page 55: Unit 5 powerpoint[1] algebra (1)

The sum of three consecutive integers is 351. Find the numbers.

Let the middle one of the three integers be n. Then the other two will be n − 1 and n + 1.

Adding these gives the equation:(n − 1) + n + (n + 1) = 351

Tidy up the equation: 3n = 351Divide by 3: n = 117The three integers are 116, 117 and 118.

(Check: 116 + 117 + 118 = 351, as expected.)

Page 56: Unit 5 powerpoint[1] algebra (1)

Nicholas buys two books for his nieces, Amy and Becky, for Christmas. Becky’s book costs £5 more than Amy’s and together the bill came to £21. How much were the books?

Page 57: Unit 5 powerpoint[1] algebra (1)

Nicholas buys two books for his nieces, Amy and Becky, for Christmas. Becky’s book costs £5 more than Amy’s and together the bill came to £21. How much were the books?

Let the price of Amy’s book be £P. Then Becky’s book cost five pounds more or £(P + 5).

The total cost was £21, giving the equation:P + (P + 5) = 21

Tidy up: 2P + 5 = 21Subtract 5: 2P = 16Divide by 2: P = 8So the books were £8 and £13.

(Check: if the books are £8 and £13, then the total cost is £21, as expected.)

Page 58: Unit 5 powerpoint[1] algebra (1)

Great Uncle Theo is planning some donations to charity. He has £1000 which he wants to divide between the Cats’ Home, the Dogs’ Home and the sanctuary for parrots. He wants to give the Dogs’ Home twice as much as the Cats’ Home. He also wants to give £100 more to the sanctuary for parrots than to the Cats’ Home. How much does he give to each?

Page 59: Unit 5 powerpoint[1] algebra (1)

Great Uncle Theo is planning some donations to charity. He has £1000 which he wants to divide between the Cats’ Home, the Dogs’ Home and the sanctuary for parrots. He wants to give the Dogs’ Home twice as much as the Cats’ Home. He also wants to give £100 more to the sanctuary for parrots than to the Cats’ Home. How much does he give to each?

Let the amount that Great Uncle Theo gives to the Cats’ Home be £c.Then he will give the Dogs’ Home twice this or £(2c), and the sanctuary for parrots will get £(c + 100).

The total of the gifts is £1000. So we can create the equation:c + 2c + (c + 100) = 1000

Tidy up: 4c + 100 = 1000Subtract 100: 4c = 900Divide by 4: c = 225

So Great Uncle Theo gives £225 to the Cats’ Home, £450 to the Dogs’ Home and £325 to the sanctuary for parrots.

(Check: £225 + £450 + £325 = £1000, as expected.)

Page 60: Unit 5 powerpoint[1] algebra (1)

There are four people travelling in a car. Bill and Ben are the same age, Molly is twice as old as Bill and Pete is twenty years older than Molly. The sum of their ages is 86. How old is Molly?

Page 61: Unit 5 powerpoint[1] algebra (1)

There are four people travelling in a car. Bill and Ben are the same age, Molly is twice as old as Bill and Pete is twenty years older than Molly. The sum of their ages is 86. How old is Molly?

Although you are asked to find Molly’s age, you may find it easier to start by letting Bill’s age be b years. Then you will avoid fractions!

If Bill’s age is b years, then Ben’s age is also b years. Molly is twice as old as Bill so she is 2b years old. Pete is 20 years older than Molly so he is (2b + 20) years old.

The sum of their ages is 86, so we can create the equation:b + b + 2b + (2b + 20) = 86

Tidy up: 6b +20 = 86Subtract 20: 6b = 66Divide by 6: b = 11

So Bill is 11 years old and from this we can work out that Molly is 22 and Pete is 42.

(Check: Adding the ages together, 11 + 11 + 22 + 42 = 86, as expected.)

Page 62: Unit 5 powerpoint[1] algebra (1)

Pirate Jack has left you instructions for finding the treasure.‘Start at the big oak tree. Take a certain number of paces to the north, then take that number plus 5 to the east. Finally, take half of the number of paces you took to the north, but this time, go in a north easterly direction. There you will find the treasure. In total you will take three times the number of paces you took to the north.’How many paces should you take to the north?

http://office.microsoft.com/en-gb/clipart

Page 63: Unit 5 powerpoint[1] algebra (1)

Pirate Jack has left you instructions for finding the treasure.‘Start at the big oak tree. Take a certain number of paces to the north, then take that number plus 5 to the east. Finally, take half of the number of paces you took to the north, but this time, go in a north easterly direction. There you will find the treasure. In total you will take three times the number of paces you took to the north.’ How many paces should you take to the north?

Let the number of paces that you take to the north be x.

Then you must take (x + 5) paces to the east and paces to the north east.

The total number of paces is 3x.Create an equation:

2

x

( 5) 32

xx x x

Tidy up: 2 5 32

xx x

Multiply by 2: 4x + 10 + x = 6xTidy up: 5x + 10 = 6xSubtract 5x: 10 = xSwap sides: x = 10So you should take 10 paces to the north.

(Check: You take 10 paces to the north, then that number plus 5, which gives 15, to the east and half the number you take to the north, which is 5, to the north east. In total, you take 10 + 15 + 5 = 30 paces which is 3 times the number of paces you took to the north.)

Page 64: Unit 5 powerpoint[1] algebra (1)

A magic square has the property that the numbers in each column, each row and each diagonal add to the same total. Substitute a = 5, b = 3 and c = −1 in the square below and check that the result is a magic square.

a + c a + b − c a − b

a − b − c a a + b + c

a + b a − b + c a − c

Now use algebra to explain why the property works for all values of a, b and c.

Page 65: Unit 5 powerpoint[1] algebra (1)

Substituting a = 5, b = 3 and c = −1 gives:

If you use algebra, you will find that the sum of each row, each column and each diagonal is 3a. For example, the sum of the first row is

4 9 2

3 5 7

8 1 6

( ) ( ) ( ) 3a c a b c a b a

Page 66: Unit 5 powerpoint[1] algebra (1)

1. In the expression (4 − a) the coefficient of a is −1.

2.

3. 3(2x − 3) = 6x − 3

4. (2a + b) − (a + b) = a

5. x = 2 is a solution of the equation 2x + 1 = 5.

6. x = 1 is a solution of the equation 2x + 1 = 7 − x.

2 3x xy x x xy