Literal calculation: answer key to 4th grade math exercises in PDF.

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Literal calculation with the answer key for 4th grade math exercises in PDF. Know how to develop an expression and double distributivity.

Exercise 1:
Write the given expressions without parentheses:
a. – (3+x)= – 3 – x
b. – (2a+4)= – 2a – 4
c. – (- 3+x)=3 – x
d. – (5 – x)= – 5+x
e. -(7-2y)= – 7+2y
f. – (-6-4x)= 6+4x

Exercise 2:
Reduce each of the following expressions:

a. $\,2x\,\times \,7=14x\,$
b. $\,-5y\,\times \,(-2)=10y\,$
c. $\,4x\,\times \,(-5)=-20x\,$
d. $\,-5\,\times \,9a\,=-45a$
e. $\,-3x\,\times \,x=\,-\,3x^2\,$
f. $\,5b\,\times \,(-\,2b)=\,-\,10b^2$
g. $\,\frac{2}{3}a\,\times \,(-\,6a)=\,-\,4a^2\,$
h. $\,3x-5+4x-13-9x=-2x-18\,$
i. $\,-2x+3-9x-4+3x=-8x-1\,$
j. $\,5x-2-4x+7-3x-2-9x-11=-11x+8\,$

Exercise 3:

Remove the parentheses and then reduce each expression.
a. $\,25-(2a-3)=25-2a+3=-2a+28\,$
b. $\,3a-(-2a+7)=3a+2a-7=5a-7\,$
c. $\,-(a+3b)+(b-2a)=-a-3b+b-2a=-3a-2b\,$
d. $\,(5+x)-(7x-5)=5+x-7x+5=-6x+10\,$
e. $\,(x^2-5x)+(2x^2+7x-8)=x^2-5x+2x^2+7x-8=3x^2+2x-8\,$
f. $\,(3x^2-5x-4)-(-4x^2+7x+5)\\=3x^2-5x-4+4x^2-7x-5\\=7x^2-12x-9\,$
g. $\,(\frac{3}{4}a^2+\,\frac{2}{3}a-4)-(\frac{1}{2}a^2+\frac{1}{3}a+3)\\=\frac{3}{4}a^2+\,\frac{2}{3}a-4-\frac{1}{2}a^2-\frac{1}{3}a-3\\=(\frac{3}{4}-\frac{1}{2})a^2+(\frac{2}{3}-\frac{1}{3})a-4-3\\=\frac{1}{4}a^2-\frac{1}{3}-7$

Exercise 4:
Expand then reduce the expressions.

a. $\,(x+3)(x+2)=x^2+2x+3x+6=x^2+5x+6$
b. $\,(2x+1)(x+4)=2x^2+8x+x+4=2x^2+9x+4\,$
c. $\,(5x+6)(2x+3)=10x^2+15x+12x+18\,$
d. $\,(7x+5)(8+9x)=56x+63x^2+40+45x=63x^2+101x+40\,$
e. $\,(x+\frac{1}{3})(x+2)=x^2+2x+\frac{1}{3}x+\frac{2}{3}=x^2+\frac{6}{3}x\,+\frac{1}{3}x+\frac{2}{3}\\=x^2+\frac{7}{3}x\,+\frac{2}{3}$
f. $\,(\frac{1}{2}x+3)(x+2)=\frac{1}{2}x^2+\frac{2}{2}x+3x+6=\frac{1}{2}x^2+4x+6$

Exercise 5:
Expand then reduce the expressions.

a. $\,(x+5)(x-3)=x^2-3x+5x-15=x^2+2x-15\,$
b. $\,(3x-7)(2+x)=6x+3x^2-14-7x=3x^2-x-14\,$
c. $\,(2x-3)(4x-1)=8x^2-2x-12x+3=8x^2-14x+3\,$
d. $\,(4x-2)(5x-3)=20x^2-12x-10x+6\,$
e. $\,(4x+1)(-4x-1)=-16x^2-4x-4x-1=-16x^2-8x-1\,$
f. $\,(x+1)^2=(x+1)(x+1)=x^2+x+x+1=x^2+2x+1\,$
g. $\,(2x-3)^2=(2x-3)(2x-3)=4x^2-6x-6x+9=4x^2-12x+9\,$

Exercise 6:

In this exercise $\,x\,$ is a number greater than 3.

We propose to express the area $\,A$ of the colored surface as a function of $\,x\,$

1.a. Explain why the area :
– of rectangle ABCD can be written $\,x(2x+1)$ ;
– of the rectangle AEFG can be written $\,x(x-3)$ .

The area of a rectangle is the product of its length and width….

b. After developing the previous literal expressions, express the area $\,A$ as a function of $\,x\,$ .

$x(2x+1)=2x^2+x\,;\,\,x(x-3)=x^2-3x$

– Show that the area $\,A$ can also be written :

It is the sum of the area of the small orange rectangle and the large orange rectangle below.

$\,A=(x+1)(x-3)+3(2x+1)$ .

or :

$\,A=A_{ABCD}-A_{AEFG}=x(2x+1)-x(x-3)$

– Expand and reduce this expression.
$\,A=(x+1)(x-3)+3(2x+1)=x^2-3x+x-3+6x+3=x^2+4x$ .

3. calculate the value of $\,A$ for $\,x\,=\,10$ .

When x = 10, the area of $\,A$ is :

$\,A=10^2+4\times \,10\,=\,100\,+40=\,140\,.$

Exercise 7:

A = x (x + 2) A=x²+2x

B = 5x (x +3) B=5x²+15x
C = 2x (3x – 5) C=6x²-10x
D = – 3x (1 – 4x) D=-3x+12x²
E = (x + 2) (-x + 3) E= -x²+3x-2x+6 E=-x²+x+6
F = (2x + 3) (4x – 1) F=8x²-2x+12x-3 F=8x²+10x-3
G = (5 – 3y) (6 – 2y) G=30-10y-18y+6y² G=6y²-28y+30

Exercise 8:

Develop and reduce.

A = (x + 3) (x – 2) + (2x + 4) (x + 5)

A=x²-2x+3x-6+2x²+10x+4x+20
A=3x²+15x+14

B = (2x – 1) ( 7x + 8) – (5 – 4x) (3x + 1)
B=14x²+16x-7x-8-(15x+5-12x²-4x)
B=14x²+16x-7x-8-15x-5+12x²+4x
B=26x²-2x-13

C = (3x + 4) ( 7x – 1) – (2x + 5) (3x – 2)
C=21x²-3x+28x-4-(6x²-4x+15x-10
C=21x²+25x-4-6x²+4x-15x+10
C=15x²+14x+6

Exercise 9:
A = (x – 3) (3x – 1) – 2x² + 4

for x = 2 :
We replace x by 2 in the expression.
A=(2-3)(3×2-1)-2×2²+4
A=-1×5-8+4
A=-5-8+4
A=-13+4
A= – 9

Exercise10:

For Figure A:

Reminder: the area of a rectangle is equal to the product of the length by the width.

A = (2x+1)(3x-2)
A = 6x²-4x+3x-2
A= 6x²-x-2

For Figure B:

Reminder: the area of a triangle is equal to half the length of the base multiplied by the height.
B= 0.5x4X(X+3)
B=2X(X+3)
B=2X²+6X

Exercise 11:

Factor by looking for a common factor.

A = 11n + 11
A= 11(n+1)

B = x ² + 5x
B= x(x+5)

C = 14t² – 21t
C=7t(2t-3)

D = (x + 5)(x + 8 ) + 2 (x + 5)
D= (x+5)(x+8+2)
D=(x+5)(x+10)

E = (2x – 9) (3x + 7) + (2x – 9) (6 – 2x)
E=(2x-9)(3x+7+6-2x)
E=(2x-9)(x+13)

F = (5x – 3) (7x – 9) – (3x + 4) (5x – 3)
F=(5x-3)[(7x-9)-(3x+4)]
F=(5x-3)(7x-9-3x-4)
F=(5x-3)(4x-13)

G = (7x + 1) ² + (7x + 1) (2x + 5)
G=(7x+1)(7x+1+2x+5)
G=(7x+1)(9x+6)

H = (2a +3) (5a – 1) – (2a +3)²
H= (2a+3) [(5a-1)-(2a+3)]
H=(2a+3)(5a-1-2a-3)
H=(2a+3)(3a-4)

Exercise 12:

Houses are represented step by step with the help of matches as shown below.

1. How many matches will be needed in steps #4 and #10? Answer without drawing a picture.

In step 4, you will need 17 matches (13 +4).

2. Check if you had found the right number

yes
3. How many matches will be needed in step #2007?

$4\times \,2007+1=8029$

It will take 8,029 matches in the 2007 stage.

4. How to express the number of matches for any step?
Let us consider a step n° with a positive relative integer.

Then there will be $4\times \,n+1=4n+1$ matches at the stage .

Exercise 13:

The teacher wrote the following exercise on the board:

Calculate

23 × 7 + 3 ;
23 × 8 + 3;
23 × 9 + 3;
23 × 10 +3
23 × 11 + 3;
23 × 12 + 3;
23 × 13 + 3;
23 × 14 + 3

A classmate is absent.

What instructions to give him on the phone, without dictating all the calculations.

The instruction is good if the classmate knows exactly what to do.

Let x be a relative integer
calculate the literal expression for x ranging from 7 to 14.

Exercise 14:

1) A hiker covers 5 km in 1 hour and 15 minutes. What is its average speed in km/h?
Justify
$5kmrightarrow\,1h15min$

or
$60minrightarrow\,\,1h$

$15minrightarrow\,\,0,25h$

$5kmrightarrow\,\,1,25h$

$\frac{5}{1,25}kmrightarrow\,\,1h$

$4kmrightarrow\,\,1h$

Conclusion: the average speed of the hiker is $4\,km/h$ .

2) A car is travelling at a speed of 50 km/h.
How long does it take to cover 110 kilometers?
Give the result in hours and minutes.

$50kmrightarrow\,1h$

$110kmrightarrow\,x\,\,h$

$x=\frac{110\times \,1}{50}=2,2h$

or
$1hrightarrow\,60min\\0,2\,hrightarrow\,x\,min$

so $x=\frac{0,2\times \,60}{1}=12\,min$

The vehicle travels 110 kilometers in 1 hour and 12 minutes.

Exercise 15:
a is a non-zero decimal number.
Give a literal expression for :
1) double the square of a .

2) the square of the double of a .

3) half the square of a :
$\frac{a^{2}}{2}$
4) the square of half of a .
$(\frac{a}{2})^2$
5) the square of the opposite of a .
$(-a)^2=(-a)\times \,(-a)=a^2$
6) the opposite of the square of a .

7) the square of the inverse of a .
$(\frac{1}{a})^2=\frac{1}{a^2}$
8) the inverse of the square of a .
$\frac{1}{a^2}$

Exercise 16:

1. Simplify the following entries:

2. expand and reduce the following expressions:

Exercise 17:

A concert hall can hold 600 people. There are x number of seats and the others are standing. Standing places cost 15€ and seats 25€.

1°) What do the expressions: a- 600 – x represent? b- 25x ? c- 15 (600 – x) ?

600-x is the number of standing places

25x is the total “price” of all the seats

15.(600-x) is the total “price” that all the standing places bring in.

2°) Express, as a function of x, the total revenue in euros if all the places are occupied.

total “price” of the room = 25x+15(600-x)

3°) Calculate this revenue if x = 200.

25×200+15.(600-200) = 25*200 + 15 * 400 (here it is clear that there are 200 seats and 400 standing places)

Total revenue = 25*200 +30*200 ( since 15*400 = 15 *2 *200 ) = 55*200 = 110*100 = 11 000€

4°) What is the number of seats if the room is full and if the revenue is 12 500 €?

To answer we don’t know exactly how many seats, so that’s our unknown x, and we’re left with an equation with one unknown.

25x+15(600-x) =12500, and x € N, x € [0,600] ( which means x natural number : example {0;1;2;…200;….;599;600} )

<==>25x-15x+15*600=12500 <==>10x+45*200=12500 (because 15*600 =15*3*200) <==>10x+90*100=12500 (because 45*200 =45*2*100)

<==> 10x +9000 =12500 and it is now that we will subtract on each side of the equal sign the value of the integer “9000”

<==> 10x+9000 -9000 = 12500 -9000

<==>10x=3500 and it is now that we will divide on each side of the equal sign by the multiple of x

<==> $\frac{10x}{{\color{Blue}\,10}}=\frac{3500}{10}$

<==> x = 350, so there were 350 seats to realize such a revenue (12 500€)

Exercise 18:

Expand the following expressions:

A = 6 (2x + 8) =12x+48

B = 7 (5x – 1) =35x-7

C = -4x (x – 9)=-4x²+36x

D = (3x + 4) (2x + 3) =6x²+9x+8x+12=6x²+17x+12

E = (7x + 5) (5x + (-3)) =35x²-21x+25x-15=35x²+4x-15

F = (2x + 9) (7x – 1)=14x²-2x+63x-9=14x²+61x-9

Exercise 19:

A calculation program is given:

Choose a number.

Add 2

Multiply the sum obtained by the chosen number

Add 1 to this product

Write the result.

1) Write the calculations to verify that if you run this program with the number – 1 , you get 0.

-1

-1+2=1

1x(-1)=-1

-1+1=0

0+1=1

2) Give the result provided by the program when the chosen number is -6

-6

-6+2=-4

-4x(-6)=24

24+1=25

3) Give the result provided by the program when the chosen number is 4

4+2=6

6×4=24

24+1=25

4) Write the expression obtained for any number a.

a+2

a(a+2)

a(a+2)+1=a²+2a+1=(a+1)²

Exercise 20:
a) Remove the parentheses and reduce the expression M

${\color{DarkRed}\,M=7x-11}$
b) Expand and reduce N and P.

${\color{DarkRed}\,N=19x-5}$

${\color{DarkRed}\,P=6x^2-x-15}$

c) Calculate N when x is equal to 3.

$N=19\times \,3-5$
${\color{DarkRed}\,N=52}$

Exercise 21:

Mr. Hamraoui: So you had chosen 5 in the beginning.

What is Mr Hamraoui’s trick?

Mr. Hamraoui has solved an equation.

By setting the number you are looking for,

We obtain for the program :

– Choose a number:

– Multiply it by -11 :

– Add 8 :

– Multiply the result by -9 :

– Add the number chosen at the beginning:

– Adds -28 :

Anna: I find 400 .

We obtain the equation :

$x=\frac{500}{100}$

${\color{DarkRed}\,x=5}$

So the number chosen was 5.

Exercise 22:

A rectangular pool measures 10 m by 5 m.

We want to build a beach around it.

However, this beach should not be too large so as not to cost the community too much, but should not be too small so as not to penalize non-bathers.

It is estimated that the beach area should be between 110 and 120 .

It is then decided to make a preliminary project of swimming pool, by noting the number designating the width of the beach.

The number then becomes what is called an unknown.

1. Calculate the area of this range in the case where , then in the case where , and finally

For l=1 :

$A=(10+2)(5+2)-5\times \,10=34m^2$

For l=2 :

$A=(10+4)(5+4)-5\times \,10=76m^2$

For l=3 :

$A=(10+6)(5+6)-5\times \,10=126m^2$

2. In each of the above cases, can the construction project be started? Why?

No, the area must be between 110 and 120 . We can say that .

3. What method can you suggest to find a satisfactory range width?

To speed up the search for this ideal width , we try to express the area of this range as a function of the number .

Two teams will work on it:

– Calculate the area of rectangle ABCD, minus the area of the pool.

– Assemble the beach elements as shown in the sketch below.

4. Find the expressions obtained by each of the two teams.

$A=(5+2l)(10+2l)-5\times \,10=50+10l+20l+4l^2-50\\A=4l^2+30l=l(4l+30)$

5. Propose a possible beach width to launch the project.

take $l=2,7\,m$

thus $A=2,7(4\times \,2,7+30)=110,16\,m^2$

and we’ve got it right $110\,m^2\leq\,\,A\leq\,\,120\,m^2$ .

Exercise 23:

Expand and reduce the following expressions:

${\color{DarkRed},A=x^2+5x+4}$

${\color{DarkRed},B=-x^2+3x+4}$

${\color{DarkRed},C=x^2+3x-4}$

${\color{DarkRed},D=x^2-5x+4}$

Exercise 24:

Remove the parentheses and then reduce the expressions:

${\color{DarkRed},E=-x-4}$

${\color{DarkRed},F=12x+3}$

Exercise 25:

Remembering that $a^2=a\times ,a$ .

expand and reduce the following expressions :

$A=(x+4)\times ,(x+4)$

${\color{DarkRed},A=x^2+8x+16}$

$B=(2x-3)\times ,(2x-3)$

${\color{DarkRed},B=4x^2-12x+9}$

Exercise 26:

Reduce the following expressions:

${\color{DarkRed},D=-3x^2-14x+3}$

${\color{DarkRed},E=4x^2+7x-1}$

${\color{DarkRed},F=-9a+2}$

Exercise 29:

1) -2

-2-3=-5

-5x(-5)=25

25:4=6,25

-2+6,25=4,25

The result for -2 is 4.25.

7-3=4

4x(-5)=-20

-20:4=-5

-5+7=2

The result for 7 is 2

2) x

x-3

-5(x-3)

-5(x-3) :4

$\frac{-5(x-3)}{4}\,+x$

$\frac{-5}{4}(x-3)\,+x$

$-1,25(x-3)\,+x$

$-1,25x-3\times \,(-1,25)\,+x$

$-1,25x+3,75\,+x$

Exercise 30:

$c)\frac{1}{3}x-5$

Exercise 31:

A = ^{x-6-5×2-30-x}

A=-5x²-36

B=12x-x²-10+x-3-8x²+1-2x

B=-9x²+11x-12

C= -3-a+b+5a-9+(-3a-5b)

C=-3-a+b+5a-9-3a-5b

C=a-4b-12

D= x²-(3x²-15x+4)+(15x²-12x-3)

D=x²-3x²+15x-4+15x²-12x-3

D=13x²+3x-7

Exercise 32:

Exercise 33:

$A=3x\,,\,B=3x^2\,,\,C=3x\,,\,D=3x+2\,,\,E=2x^2\,,\,F=x^2+x$

$G=0\,,\,H=1+2x\,,\,I=x\,,\,J=30x^2\,,\,K=30x\,,\,L=x^2+x$

The answer key to the math exercises on literal calculation in 4th grade.

After consulting the answers to these exercises on literal calculation in the eighth grade, you can return to the exercises in the eighth grade.

Third grade exercises.

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