Equations

1. DĂ©finition of an equation :

DĂ©finition of an equation :

â–¶ An EQUATION is an equality of two expressions (the MEMBERS of the equation) in which letters
appear to represent unknown numbers. 
â–¶ These letters are called the VARIABLES of the equation.
â–¶ If we replace these variables with any random value, the equality will always be false.
â–¶ In cases where the equality is verified, the value is said to be a SOLUTION of the equation.

Exemple :

➤ t is a rational number,
➤ $(-3) \mathbf{t}+\frac{1}{4}=12+\frac{5}{11} \mathbf{t}$ is an equation; $\mathbf{t}$ is the unknown,
➤ $\left((-3) t+\frac{1}{4}\right)$ and $\left(12+\frac{5}{11} t\right)$ are the members of this equation
➤ If we replace $t$ by 0 and calculate each member of the equation separately:
$(-3) \times 0+\frac{1}{4}=\frac{1}{4} \quad 12+\frac{5}{11} \times 0=12$
– Since $\frac{1}{4} \neq 12$ So $\mathbf{0}$ is not a solution to the equation.
➤ If we replace $\mathbf{t}$ by $\left(\frac{-517}{152}\right.$ and calculate each member of the equation separately:
$\begin{aligned} & (-3) \times)+\frac{1}{4}=\frac{1551}{152}+\frac{38}{152}=\frac{1589}{152} \\ & \left.12+\frac{5}{11} \times\right)=12-\frac{2585}{1672}=\frac{20064}{1672}-\frac{2585}{1672}=\frac{17479}{1672}=\frac{1589}{152}\end{aligned}$
– Since $\frac{1589}{152}=\frac{1589}{152}$, both members are equal.
– So $\left(\frac{-517}{152}\right.$ is a solution to the equation.

2. The degree of an equation :

DĂ©finition :

â–¶ x is a rational number, $\mathrm{a} ; \mathrm{b} ; \mathrm{c}$ and d are known rational number.
â–¶ Any equation of the form $\mathbf{a x + b = 0}$ or $\mathbf{a x}=\mathbf{b}$ or $\mathbf{a x + b}=\mathbf{c x}+\mathbf{d}$ is a first-degree equation with a single unknown $x$
â–¶ The value of the number x that verifies this equation is the solution of this equation

Remark :

▸ $x$ and $y$ are two rational numbers; $2 x+\frac{1}{-4} y-1=0$ : is a first-degree equation with two variables x and y .
â–¸ x is rational number; $\mathrm{x}^2-\frac{1}{4}=0$ : is a second-degree equation with a single variable x .
â–¸ $\mathbf{x}$ is rational number; $\frac{x}{2}-\frac{1}{3}=\frac{2 x-3}{6}$ is a first-degree equation with one variable $\mathbf{x}$

DĂ©finition :

– Solving the first-degree equation with a single variable :
â–¶ It means finding the value of the variable that makes the equality true;
â–¶ And to solve this equation, we apply one of the following properties

Propriety :

– $a, b$ and $c$ are rational numbers with $c \neq 0$ :
We have: if $a=b$ Thena $\times c=b \times c$
$$
\text { if } a=b \text { Then } \frac{a}{c}=\frac{b}{c}
$$

Exemple :

$$
\begin{array}{llll}
\text { We have: } & -17=-17 & \text { We have : } & x+3=-7 \\
\text { So : } & -17+29=-17+29 & \text { So : } & x+3+(-3)=-7+(-3) \\
\text { Then : } & 12=12 & \text { Then : } & x=-10
\end{array}
$$

Propriety :

– $a, b$ and $c$ are rational numbers with $c \neq 0$ :
We have: if $a=b$ Thena $\times c=b \times c$
if $a=b$ Then $\frac{a}{c}=\frac{b}{c}$

Exemple :

$$
\begin{array}{ll}
 \\
\text { We have : } 11=11 & \text { We have : } 5 \mathrm{x}=-15 \\
\text { So : } \quad 11 \times(-4)=11 \times(-4) & \text { So : } \quad 5 x \times \frac{1}{5}=-15 \times \frac{1}{5} \\
\text { Then : } \quad-44=-44 \quad & \text { Then : } \quad x=-3
\end{array}
$$

Exemple 2 :

– x is a rational number :
– Solve the following equation : $5(x-1)-x=4(x+2)$
– We have : $\quad 5(x-1)-x=4(x+2)$
– then : $5 x-5-x=4 x+8$
– so : $4 x=4 x+8$
– Means that : $\quad 0 x=8$
– Since there is no rational number verifying this equality.
– We say that the equation has no solution

Exemple 3 :

– x is a rational number :
– Solve the following equation : $\frac{x}{2}-\frac{1}{3}=\frac{2 x-3}{6}$
– We have : $\quad \frac{\frac{x}{2}-\frac{1}{3}}{3 x-2}=\frac{2 x-3}{6}$
then : $\quad \frac{3 x-3}{6}=\frac{2 x-3}{6}$
– so : $\quad 3 x-2 x=-3+2$
– Means that : $x=(-1)$
– Since $(-1)$ is a rational number then the solution to this equation is $(-1)$

Exemple 4 :

–  x is a rational number:
– Solve the following equation : $3(x-1)+x=4 x-3$
– We have: $3(x-1)+x=4 x-3$
– So : $3 x-3+x=4 x-3$
– Means that : $4 x-3=4 x-3$
– then : $4 x-4 x=-3+3$
– Finally : $0 x=0$
– Since all the rational number can verify this equation,
– So all the rational numbers are solution to this equation

Propriety :

– x is an unknown rational number
– a and b are known rational numbers;
– A $1^{\text {st }}$ degree equation with one variable is an equality that after simplification we can write it in the form of $\mathbf{a x}+\mathbf{b}=\mathbf{0}$ where $a$ and $b$ are known numbers
– if $\mathbf{a} \neq \mathbf{0} \quad$ Then : the solution to the equation is $\frac{(-b)}{a}$
– if $\mathbf{a = 0}$ and $\mathbf{b} \neq \mathbf{0}$ Then : the equation has no solution
if $\mathbf{a = 0}$ and $\mathbf{b = 0}$ Then : All the rational numbers are solution to the equation,

Remark :

â–¸ To solve a second-degree equation with one variable, we use factorization and we write it as the form $(a x+b)(c x+d)=0$ or $(a x+b)^2=0$ then we use the following proprieties:

Propriety :

– a and b are rational numbers;
if $a \times b=0 \quad$ Then $a=0$ or $b=0$

Propriety :

– a is a rational number;
if $a^2=0 \quad$ Then $a=0$

Exemple :

– x is a rational number :
– Solve the following equation : $(3 x-6)(x+5)=0$
– We have : $\quad(3 x-6)(x+5)=0$
– So : $\quad 3 x-6=0 \quad$ or $\quad x+5=0$
– Means that: $\quad 3 x=6 \quad$ or $\quad x=-5$
– then : $\quad x=2 \quad$ or $\quad x=-5$
– Since 2 and $(-5)$ are rational numbers
– So 2 and $(-5)$ are solutions to this equation

Exemple 2 :

– x is a rational number :
– Solve the following equation : $x^2=16$
– We have : $x^2-4^2=0$
– Means that : $(x-4)(x+4)=0$
– then : $x-4=0 \quad$ or $\quad x+4=0$
– So : $x=4 \quad$ or $\quad x=-4$
– Since 4 and (-4) are rational numbers
– So 4 and (-4) are solutions to this equation

Exemple 3 :

– x is a rational number :
– Solve the following equation : $4 x^2-12 x+9=0$
– We have: $4 x^2-12 x+9=0$
– So : $(2 x)^2-2 \times 2 x \times 3+(3)^2=0$
– Means that : $(2 x-3)^2=0$
– then : $2 x-3=0$
– So : $2 x=3$
– Finally, $x=\frac{3}{2}$
– Since $\frac{3}{2}$ is a rational number
– So $\frac{3}{2}$ is solution to this equation

Remark :

â–¸ Many problems can be translated into algebraic equations. To solve problems using algebra, we follow these steps :

Step 1 : Decide the unknown quantity and allocate it a variable such as $x$.
Step 2 : Translate the problem into an equation
Step 3 : Solve the equation by isolating the variable.
Step 4 : Check that the solution satisfies the original problem and write the answer in sentence form, describing how the solution relates to the original problem.

Exemple :

– GHALI bought a calculator and a book. The price of the book is the double of the price of the calculator. Knowing that GHALI has paid 45dh in the total determine the price of each article.
Step 1: Decide the variable
– We represent the price of the calculator with the variable $x$
– So, the price of the book is represented by $2 \times x=2 x$
Step 2 : Translate the problem into an equation
– The equation that represent the problem is : $x+2 x=45$
So $3 x=45$
Step 3 : Solve the equation
– we have $3 x=45$
So, $x=45 / 3=15$
Step 4 : Check the solution
– we have the price of the calculator is 15 dh and the price of the book is 30 dh , and $15+30=45$

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