Equations and problems: answer key to 3rd grade math exercises in PDF.

The answer key to the math exercises in 3ème on solving first degree equations with one unknown. Know how to solve an equation and determine its solution set in third grade.

Exercise 1:

Exercise 2:

Determine three consecutive positive integers whose sum of squares is equal to 1325.

For ease of calculation, the following consecutive numbers will be chosen:

n-1 ; n ; n +1 with n

1- Calculation of the squares of numbers n-1 and n+1

(n-1)² = n² -2n +1

(n+1)²= n² +2n +1

2- Calculate the sum of the squares of the three numbers.

(n-1)² + n² + (n+1)² = 3n² + 2

3- Since this sum is equal to 1325, we have to solve the equation :

3n² +2 = 1325

or

3n² = 1323

n² = 441

n= = 21

Verification:

n-1 = 20

n = 21

n+1 =22

20² +21² +22² = 400 +441 + 484 = 1325

So the consecutive numbers to determine are: 20; 21; 22

Exercise 3:

Three clubs meet in a competition. Club A won one third of the medals, Club B two seventh of the medals and Club C sixteen medals. How many medals were distributed in total?

That is : the number of medals won .

Club A and B won: medals.

Club C won of the medals, i.e. 16 medals.

so

Conclusion: there were 42 medals distributed.

Exercise 4:

Solve the following equations:

1. (x-7)²-(2x+5)²=0

Property: A product of factors is zero if and only if at least one of the factors is zero.

2. (7x+1)²-(3x+4)²=0

3. (6x-1)²-(2x+1)²=0

Exercise 5:

a piggy bank contains 65 euros in coins of 1 € and 2 € in total of 35 coins.

How many 1 euro coins are there and how many 2 euro coins?

Let x be the number of coins of 1 € and y the number of coins of 2 €.

x+y=35 (1)

x+2y=65 (2)

By subtracting (2)-(1) :

x+2y-x-y=65-35

y=30

There are 30 coins of 2 € and 5 coins of 1 €.

Exercise 6:
a. 3x + 2 = 14
3x=14-2
3x=12
x=12:3
x=4
so S={ 4 }

b. 3x – 4 = 2x + 9
3x-2x=9+4
x=13

c. 5x – 4 = 8 – 3x
5x+3x=8+4
8x=12
x=12:8=3:2
therefore S={ 1,5 }

d. 3 – (5 – x) = 3 – 4x
3-5+x=3-4x
-2+x=3-4x
x+4x=3+2
5x=5
x=5:5=1
so S={ 1 }

e. 2x + 5 = 3x – 1
2x-3x=-1-5
-x=-6
x=6
so S={ 6 }

f. 2(5 – 3x) = 6(2x + 1)
10-6x=12x+6
-6x-12x=6-10
-18x=-4
x=4:18=2:9
so S={ 2:9 }

g. 4(3x – 2) – 10x = 3x – 1
12x-8-10x=3x-1
2x-8=3x-1
2x-3x=8-1
-x=7
x=-7
so S={ -7 }

h. 3(x + 2) – (x – 3) = x – 5 – 3(x + 1) + 4x
3x+6-x+3=x-5-3x-3+4x
2x+9=2x+2
0x=2-9
0x=-7
0=-7
absurd so no solution.

i. 5x + 7 = -5 + 11x
5x-11x=-5-7
-6x=-12
x=12:6=2
so S={ 2 }

j. 2x + 1 = 4(x – 2) + x
2x+1=4x-8+x
2x-4x-x=-1+8
-3x=7
x=-7:3
so S={ -7:3 }.

Exercise 7:

Exercise 11:

Solve the following equations:

1)

2)

3)

4)

5)

6)

Exercise 12:

Let ABCD be a square of dimension 10 cm . N is a point of [AD] and R is a point of [DC] such that AN is equal to DR is equal to x (in cm).
We wish to find the position of the point N for which the area of the rectangle NORTH is maximal.
1) Give a frame for x.

2)a) Express the area of NORTH as a function of x.

b) Show that the area is equal to : 25-(x-5)².

3)a) Determine the value of x for which the NORTH area is maximum where the point N is then located.

The area is maximum when the quantity is minimum, i.e. when .

N is therefore located in the middle of [AD].

b) In this case what can be said about the NORTH rectangle.

In this case NORTH is a square.

Exercise 13:

a. 3x + 2 = 14

3x=14-2
3x=12
x=12:3
x=4
so S={ 4 }

b. 3x – 4 = 2x + 9
3x-2x=9+4
x=13

c. 5x – 4 = 8 – 3x
5x+3x=8+4
8x=12
x=12:8=3:2
therefore S={ 1,5 }

d. 3 – (5 – x) = 3 – 4x
3-5+x=3-4x
-2+x=3-4x
x+4x=3+2
5x=5
x=5:5=1
so S={ 1 }

e. 2x + 5 = 3x – 1
2x-3x=-1-5
-x=-6
x=6
so S={ 6 }

f. 2(5 – 3x) = 6(2x + 1)
10-6x=12x+6
-6x-12x=6-10
-18x=-4
x=4:18=2:9
so S={ 2:9 }

g. 4(3x – 2) – 10x = 3x – 1
12x-8-10x=3x-1
2x-8=3x-1
2x-3x=8-1
-x=7
x=-7
so S={ -7 }

h. 3(x + 2) – (x – 3) = x – 5 – 3(x + 1) + 4x
3x+6-x+3=x-5-3x-3+4x
2x+9=2x+2
0x=2-9
0x=-7
0=-7
absurd so no solution.

i. 5x + 7 = -5 + 11x
5x-11x=-5-7
-6x=-12
x=12:6=2
so S={ 2 }

j. 2x + 1 = 4(x – 2) + x
2x+1=4x-8+x
2x-4x-x=-1+8
-3x=7
x=-7:3
so S={ -7:3 }

Exercise 14:

By multiplying by 6 the two members of the equation,
we obtain :

Exercise 15:
1. (x + 5)(x – 3) = 0
It is a product equation, so we use the rule :
“a product of factors is zero if and only if at least one of the factors is zero”
thus:
x+5 = 0 or x-3 = 0
x= -5 or x=3
so S = {-5 ; 3}

2. ( 2x + 7 )( -5x + 2 ) =0
We use the rule.
2x+7=0 or -5x+2=0
2x=-7 or -5x=-2
x=-7:2 or x = -2:(-5)=2:5
so S = {-3.5; 0.4}

3. 64x² – 81 = 0
it was necessary to recognize the remarkable identity.
(8x)²-9²=0
(8x+9)(8x-9)=0
we use the rule.
8x+9=0 or 8x-9 = 0
x= -9:8 or x= 9:8
so S= { -9:8 ; 9:8}

5. ( 3 – x )(2x + 7 )(-5 + x) = 0
We use the rule which is valid for two factors also for n factors.
3-x=0 or 2x+7 = 0 or -5+x=0
x=3 or x=-7:2 or x=5
therefore S={3 ; -3,5 ; 5 }

6. 49X² – 42X + 9 = 0
it was necessary to recognize the remarkable identity.
(7X)²-2x7Xx3+3²=0
(7X-3)² = 0
(7x-3(7x-3)=0
We use the rule.
7x-3=0
x=3:7
so S ={3:7}
Exercise 16:
Find the equations that have ( 2 ) as their solution:

1. 2x + 4 = 0
let’s replace x by -2 in the equation.
2x+4=2x(-2)+4=-4+4=0
so -2 verifies the equality so -2 is a solution of the equation.
2. 2x =- 4
let’s replace x by -2 in the equation.
-2x=-2x(-2)=+4-4
therefore -2 does not verify the equality therefore -2 is not a solution of the equation.

3. 6x + 2 = 10
6x+2=6x(-2)+2=-12+2=-10
so -2 verifies the equality so -2 is a solution of the equation.

4. 5x + 4 = 2x+3
let’s replace x by -2 in the equation.
-5x+4=-5x(-2)+4=10+4=14 and 2x+3=2x(-2)+3=-4+3=-1
14 gold-1
therefore -2 does not verify the equality therefore -2 is not a solution of the equation.
Exercise 17:
1. Choice of the unknown :
Let us note x :the age of Julie.

2. Mathematical translation of the statement:
his mother was 30 years old so the age of his mother is: x+30;
his brother was 4 years old so his brother’s age is: x+4

3. Putting the problem into an equation :
Julie, her brother and her mother total a century
therefore x+ (x+30) + (x+4) = 100
x+x+30+x+4=100
3x+34=100
3x=100-34
3x=66
x=66:3
x=22

Conclusion: Julie is 22, her mother is 52, and her brother is 26.

Exercise 18:

Find the length x.

In the triangles AEF and ACB ,

according to the direct part of Thales’ theorem :

Exercise 19:

Solve the following equations after factoring using a remarkable identity: a) x² +14x+49=0

b) y²-12y+36=0

c) 4x²-20x +25=0

d) 24z+16+9z²=0

Exercise 20:

1) Factorize E = 4x²-49=(2x+7)(2x-7)

2) Let the expression F= (2x-7)(-5x+9)+4x²-49 .

a) expand then reduce F.

F= -10x² + 18x + 35x – 63 + 4x²- 49

F = – 6x²+53x-112

b) calculate the exact value of F when , , .

c) write F as a product of factors of the first degree.

d)Solve the equation F=0 .

A product of factors is zero if and only if at least one of the factors is zero.

Exercise 21:

We give the expression A= (2x-3)²-(4x+7)(2x-3) .

1. Develop and reduce A.

2. Factorize A .

3. Solve the equation (2x-3)(-2x-10)= 0

Exercise 22:

A baker sells two thirds of his baguettes in the morning.

In the afternoon, he sells another 90.

In the evening, he has 20 chopsticks left.

How many chopsticks had he baked for the day?

Let’s note x: the number of baguettes prepared for the day.

He prepared 330 chopsticks.

Exercise 23:

The following two questions are related.

1) Develop .

2) Solve the equation .

Ownership:

A product of factors is zero if at least one of the factors is zero.

Exercise 24:

The lines (FC) and (DA) are parallel if .

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