Equations: 3rd grade math course download PDF.

A course on solving the first degreeequation with one unknown and studying problems leading to an equation is timely for student development. We will see the different properties of solving an equation.

The student will need to know the definition of an equation and how to use the various properties to solve it. It must also solve product equations and equations of the type x²=a (where a is a relative number). We will end this chapter with real life problems in the third grade.

I. The first degree equation with one unknown

Ownership:

An equality is not changed by adding (or subtracting) the same number to each member of the equation.

An equality is only modified if each member of the equation is multiplied (or divided) by the same non-zero number.

Example:

Solve the equation 7x+2=4x+9.

7x+2-4x=4x+9-4x

3x+2=9

3x+2-2=9-2

3x=7

x=\frac{7}{3}

The solution of this equation is x=\frac{7}{3}.

II. Product equations

Ownership:

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

A\times  \,B=0 is equivalent to A\,=\,0 or B\,=\,0.

Examples:

Solve the following equations:

1.(x-3)(2x+6)=0

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

Therefore:

x-3=0\\x=3 or 2x+6=0\\2x=-6\\x=-\frac{6}{2}=-3

The two solutions of this equation are x=3 and x= – 3.

2.(x+1)(2x-7)+(x+1)(3x+2)=0\\(x+1)%5B(2x-7)\,+(3x+2)%5D=0\\(x+1)(5x-5)=0

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

Therefore:

x+1=0\\x=-1 or 5x-5=0\\5x=5\\x=1

The two solutions of this equation are x = -1 and x = 1.

III. Solving problems and equations

Exercise #1:

Find 3 consecutive integers whose sum is equal to 984.

The smallest number will be considered as unknown.

We note x the smallest number then :

x,+x+1+x+2=984\\3x+3=984\\3x=984-3\\3x=981\\x=\frac{981}{3}\\x=327

The three numbers we are looking for are 327, 328 and 329.

Exercise #2:

A sports club offers the following formula: a membership card for 12 € then

the use of the gymnasium charged at 4,50 € per hour.

Let us denote by x the number of hours of use of the gymnasium.

Determine the price to pay according to the number of hours of use.

After how many hours of use the price is 79,50 €?

We have:

12+4,5x=79,5\\4,5x=79,5-12\\4,5x=67,5\\x=\frac{67,5}{4,5}\\x=15

At the end of the fifteenth hour, the price to pay will be 79.50 euros.

Exercise #3:

The tank of a car is one third full. We add 42 liters to fill it.

What is its capacity?

The total capacity of the tank will be chosen as the unknown.

Let x be the capacity in liters of this tank.

\frac{1}{3}x+42=x\\\frac{1}{3}x-x=-42\\-\frac{2}{3}x=-42\\\frac{2}{3}x=42\\x=\frac{42\times  \,3}{2}=63\,L

This tank has a capacity of 63 liters.

Cette publication est également disponible en : Français (French) العربية (Arabic)

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