To solve a first-degree equation or inequality, it is possible to use various general methods (including the balancing equation method, the inverse operation method, the hidden term method, and trial-and-error). These methods are explained in the following concept sheet:
A first-degree equation is one that can be reduced to the form |0 = ax + b|.
When we solve such an equation, we attempt to determine the value of the variable that makes the equation true. To do this, it is important to remember that to keep the equality in the equation, we must apply the same calculations to both sides of the equation.
What is the value of |x| in the equation below? ||2x + 3 = 7||
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To eliminate the term |+3| from the left, we must subtract |3| from both sides of the equation. ||\begin{align}2x + 3 \color{red}{-3}& = 7 \color{red}{-3}\\ 2x &= 4\end{align}||
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We try to find the value of a single |x|. To do this, we divide each side of the equality by |2.| ||\begin{align}\displaystyle \frac{2x}{\color{red}{2}}&=\frac{4}{\color{red}{2}}\\ x &= 2\end{align}||
Answer: The value of |x| is |2.|
What is the value of |x| in the equation below? ||\dfrac{2x}{3} - 16 = -6||
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To eliminate the term |-16| from the left, we have to add |16| to both sides of the equation. ||\begin{align}\displaystyle \frac{2x}{3} - 16 \color{red}{+ 16} &= -6 \color{red}{+ 16}\\ \displaystyle \frac{2x}{3} &= 10\end{align}||
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We need to isolate the |2x|. Thus, we must multiply both sides of the equation by |3.| ||\begin{align}\displaystyle \frac{2x}{3}\times \color{red}{3}& = 10\times \color{red}{3}\\ 2x &= 30\end{align}||
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To isolate the |x|, we divide both sides of the equation by |2.| ||\begin{align}\displaystyle \frac{2x}{\color{red}{2}} &= \frac{30}{\color{red}{2}}\\ x &= 15\end{align}||
Answer: The value of |x| is |15.|
What is the value of |x| in the equation below? ||\displaystyle -2(x-9)=\frac{7}{3}||
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We distribute |-2| to all the terms in brackets. ||\displaystyle -2x+18=\frac{7}{3}||
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We move all the constant terms to one side of the equality. ||\begin{align}\displaystyle -2x+18\color{red}{-18}&=\frac{7}{3}\color{red}{-18}\\ -2x&=-\frac{47}{3}\end{align}||
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We divide both sides of the equality by |-2.| ||\begin{align}\displaystyle -2x\color{red}{\div -2}&=-\frac{47}{3}\color{red}{\div -2}\\ x&=\frac{47}{6}\end{align}||
Note: Since the division results in an infinite (or repeating) decimal, it’s better to leave the answer as a fraction.
Answer: The value of |x| is |\displaystyle \frac{47}{6}.|
When an equation has several fractions with different denominators, we can put all the terms over a common denominator. Then, we get rid of the denominator and keep only the numerators of each term. This trick is explained in the example below.
What is the value of |x| in the equation below? ||\displaystyle \frac{8}{3}x+1=\frac{5}{9}x-\frac{1}{4}||
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We put all the terms over a common denominator. Let’s use |36.| ||\begin{align} \displaystyle \frac{8\color{blue}{\times 12}}{3\color{blue}{\times 12}}x+\frac{1\color{blue}{\times 36}}{1\color{blue}{\times 36}}&=\frac{5\color{blue}{\times 4}}{9\color{blue}{\times 4}}x-\frac{1\color{blue}{\times 9}}{4\color{blue}{\times 9}}\\ \\ \frac{96}{36}x+\frac{36}{36}&=\frac{20}{36}x-\frac{9}{36}\end{align}||
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Since all the terms have a common denominator, we can simplify the equation by multiplying each term by 36 and keeping only the numerators. ||96x+36=20x-9||
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We group together all the terms containing the variable |x| on one side of the equal sign. ||\begin{align}96x+36\color{red}{-20x}&=20x-9\color{red}{-20x}\\ 76x+36&=-9\end{align}||
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We group all the constant terms on the other side of the equality. ||\begin{align}76x+36\color{red}{-36}&=-9\color{red}{-36}\\ 76x&=-45\end{align}||
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We divide both sides of the equality by |76.| ||\begin{align} \displaystyle \frac{76x}{\color{red}{76}}&=\frac{-45}{\color{red}{76}}\\ x&=-\frac{45}{76}\end{align}||
Answer: The value of |x| is |\displaystyle -\frac{45}{76}.|
Dans la MiniRécup suivante, tu auras accès à une vidéo interactive où on approfondit la résolution d'équations dans des problèmes en contexte.
Like first-degree equations, first-degree inequalities have only one variable. To solve an inequality, we proceed in much the same way as solving an equation: we isolate the desired variable. The difference between equations and inequalities is that inequalities have an inequality sign instead of an equals sign.
To solve an inequality, we must remember to apply the same calculations to both sides of the inequality.
The inequality is: ||2x + 5 \le 9||
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We group all the constant terms on one side of the inequality. ||\begin{align}2x + 5 \color{red}{- 5} &\le 9 \color{red}{- 5}\\ 2x &\le 4\end{align}||
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We divide both sides of the inequality by |2.| ||\begin{align}\displaystyle \frac{2x}{\color{red}{2}} &\le \frac{4}{\color{red}{2}}\\ x& \le 2\end{align}||
Answer: |x| is less than or equal to |2.|
When we divide or multiply by a negative number, we must flip or reverse the inequality sign.
The inequality is: ||10 - 2x > 3x + 15||
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We group together the terms containing the variable |x| on one side of the inequality. ||\begin{align}10 - 2x \color{red}{- 3x} &> 3x + 15 \color{red}{- 3x}\\ 10 - 5x &> 15\end{align}||
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We group all the constant terms on the other side of the inequality. ||\begin{align}10 - 5x \color{red}{- 10} &> 15 \color{red}{- 10}\\ -5x &> 5\end{align}||
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We divide both sides of the inequality by |-5.| ||\begin{align}\displaystyle \frac{-5x}{\color{red}{-5}} &> \frac{5}{\color{red}{-5}}\\ x &\color{blue}{<} -1\end{align}||Note: Since there is a division by a negative number, the inequality sign is reversed.
Answer: The value of |x| must be smaller than |-1.|
The inequality is: ||\displaystyle \frac{-11x + 15}{3} < 6 - 4x||
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We eliminate the denominator on the left side of the inequality. To do this, we multiply both sides of the inequality by |3.| ||\begin{align}\displaystyle \frac{-11x + 15}{3}\times \color{red}{3} &< (6 - 4x)\times \color{red}{3}\\ -11x + 15 &< 18 - 12x\end{align}||
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We group together the terms containing the variable |x| on one side of the inequality. ||\begin{align}-11x + 15 \color{red}{+ 12x}& < 18 - 12x \color{red}{+ 12x}\\ x + 15 &< 18\end{align}||
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We group together all the constant terms on the other side of the inequality. ||\begin{align}x + 15 \color{red}{- 15} &< 18 \color{red}{- 15}\\ x& < 3\end{align}||
Answer: The value of |x| must be smaller than |3.|
We can represent the set of solutions to a first-degree inequality in multiple ways: