Inequations: answer key to 3rd grade math exercises in PDF.

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Answers to math exercises in 9th grade on solving inequalities. Solve an inequation and represent the solution set on a graduated line in third grade.

Exercise 1:

Solve the following inequations:

a. 2x-7>\,2-x

2x+x>\,7+2

3x>\,9

x>\,\frac{9}{3}

{\color{DarkRed}\,x>3}

b. 5+y\geq\,\,7+3y

y-3y\geq\,\,7-5

-2y\geq\,\,2

y\leq\,\,-\frac{2}{2}

{\color{DarkRed}\,y\leq\,\,-1}

c. 3t+2\leq\,\,2(1+3t)

3t+2\leq\,\,2+6t

2-2\leq\,\,6t-3t

0\leq\,\,3t

{\color{DarkRed}\,0\leq\,\,t}

Exercise 2:

Solve the following inequalities:

1) 2x+2\geq\,\,,4x+6.

2x-4x\geq\,\,6-2

-2x\geq\,\,4

x\geq\,\,-\frac{4}{2}

{\color{DarkRed}\,x\geq\,\,-2}

2) 2(3x+3)>\,5(2x+3)

6x+6>\,10x+15

6x-10x>\,15-6

-4x>\,9

{\color{DarkRed}\,x>-\frac{9}{4}\,}

3) 3(2x-6)<\,2x+6

6x-18<\,2x+6

6x-2x<\,6+18

4x<\,24

x<\,\frac{24}{4}

{\color{DarkRed}\,x<\,6}

4) \frac{x+3}{3}\geq\,\,2

x+3\geq\,\,6

x\geq\,\,6-3

{\color{DarkRed}\,x\geq\,\,3}

5) \frac{2x-3}{3}\leq\,\,-5

2x-3\leq\,\,-15

2x\leq\,\,-15+3

2x\leq\,\,-12

x\leq\,\,-\frac{12}{2}

{\color{DarkRed}\,x\leq\,\,-6}

6) \frac{3-4x}{5}>\,1

3-4x>\,5

-4x>\,5-3

-4x>2

x>-\frac{2}{4}

{\color{DarkRed}\,x>-\frac{1}{2}}

Exercise 3:
Let L be the length of this rectangle.
L > 5,3 cm
The perimeter of this rectangle is equal to : 2L + 2 × 5.3 = 2L + 10.6

2L + 10,6 ≤ 37

2L + 10,6 – 10,6 ≤ 37 – 10,6

2L ≤ 26,4

\frac{2L\,}{2}\leq\,\,\frac{26,4\,}{2}

L ≤ 13,2

Conclusion: the length of this rectangle is between 5.3 cm and 13.2 cm.

Exercise 4:

Let x be the largest of three consecutive integers.

The previous one is equal to x – 1 and the smallest one is equal to x – 2.

The sum of these three integers is equal to : x – 2 + x – 1 + x = 3x – 3

12 < 3x – 3< 27

12 + 3 < 3x – 3 + 3 < 27 + 3

15 < 3x < 30

\frac{15}{3}\,<\,\frac{3x}{3}\,<\,\frac{30}{3}

5 < x < 10

The largest of these three integers is 6 , 7, 8 or 9.

Exercise 5:

Represent on a graduated line the solutions of the inequation :

-2x+7 < 5x + 29.

-2x-5x<\,29-7

-7x<\,22

x>\,-\frac{22}{7}

Conclusion: {\color{DarkRed}\,S=%5D-\frac{22}{7}\,;+\infty%5B}

Exercise 6:

One amusement park offers two membership options:

Formula A: The annual card costs 55 € and the price of an entry is 20 €.

Formula B: The annual card costs 80 € and the price of an entry is 15 €.

We note y the number of entries .

1. Express, as a function of y, the cost per year with the formula A .

A=55+20y

2. Express, as a function of y, the cost per year with the formula B .

B=80+15y

3. From how many entries in the year, does formula B prove to be

the most interesting?

B>\,A

80+15y>55+20y

15y-20y>55-80

-5y>-25

y<\,\frac{-25}{-5}

y<\,5

Formula B is the most interesting for less than 4 entries.

Landscape and inequality

Exercise 7:

Consider the inequation:

x^2+3x\geq\,\,(x-1)(x+2)

1. Justify that 0 is a solution of this inequation.

0^2+3\times  \,0=0\,et\,\,(0-1)(0+2)=-2

or 0\geq\,-2

so 0 is the solution of this inequation.

2. -\frac{1}{2} is it a solution of this inequation?

(-\frac{1}{2})^2+3\times  \,(-\frac{1}{2})\,\,\,et\,\,\,((-\frac{1}{2})-1)((-\frac{1}{2})+2)

\frac{1}{4}-\frac{3}{2}\,\,\,et\,\,\,(-\frac{1}{2}-\frac{2}{2})(-\frac{1}{2}+\frac{4}{2})

\frac{1}{4}-\frac{3}{2}\,\,\,et\,\,\,-\frac{3}{2}\times  \,\frac{3}{2}

-\frac{5}{4}\,\,\,et\,\,\,-\frac{9}{4}

or -\frac{5}{4}\,\geq\,\,-\frac{9}{4}

so it is a solution of this inequation.

3. After expanding the second member, solve this inequation.

x^2+3x\geq\,\,(x-1)(x+2)

x^2+3x\geq\,\,x^2+2x-x-2

x^2+3x\geq\,\,x^2+x-2

x^2+3x\,-x^2-x+2\geq\,\,0

2x+2\geq\,\,0

2x\geq\,-2

{\color{DarkRed}\,x\geq\,-1}

The solutions are all numbers greater than -1.

Remark:

We verify that the two values

of the first questions are well overall solution.

Answers to math exercises on inequalities in 3rd grade

After having consulted the answers to these math exercises on solving inequalities in 3rd grade, you can go back to the exercises in 3rd grade

Third grade exercises.

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