Solving equations is the process of determining the value(s) of the unknown variables that make an equation true.
There are different methods for solving an equation:
Balancing equations consists of isolating a variable to one side of the equation, using the rules for transforming equations.
Like the scales of a balance in a state of equilibrium, the rules for balancing equations allow us to transform the equation while keeping both sides of the equation equal.
Find the value of |x| in the following equation: |2x + 5 = x + 7|
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To eliminate the algebraic term |x| from the right-hand side, subtract it from both sides of the equation. ||\begin{align}2x + 5 \color{red}{- x} &= x + 7 \color{red}{- x} \\ x + 5 &= 7 \end{align}||
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To isolate |x| on the left-hand side, subtract |5| from both sides of the equation. ||\begin{align} x + 5 \color{red}{- 5} &= 7 \color{red}{- 5} \\ x &= 2 \end{align}||
Conclusion: |x = 2.|
Find the value of |x| in the following equation: |\displaystyle \frac{3x}{4} - 2{.}5 = 2{.}3|
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To isolate the term |\displaystyle \frac{3x}{4}| in the left-hand side, add |2{.}5| to both sides of the equation. ||\begin{align} \frac{3x}{4} - 2{.}5 \color{red}{+ 2{.}5} &= 2{.}3 \color{red}{+ 2{.}5} \\ \frac{3x}{4} &= 4{.}8 \end{align}||
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To isolate the term |3x| on the left-hand side, multiply both sides of the equation by |4|. ||\begin{align} \frac{3x}{4}\color{red}{\times 4} &= 4{.}8\color{red}{\times 4} \\ 3x &= 19{.}2 \end{align}||
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To isolate |x| on the left-hand side, divide both sides of the equation by |3|. ||\begin{align} \color{red}{\frac{\color{black}{3x}}{3}} &= \color{red}{\frac{\color{black}{19{.}2}}{3}} \\ x &= 6{.}4 \end{align}||
Conclusion: |x = 6{.}4.|
The inverse operations method consists of isolating an unknown variable by applying the inverse of each operation to both sides of the equation.
When using the inverse operations method, apply the inverse of each operation in the reverse order of the order of operations.
Find the value of |x| in the following equation: |\displaystyle \frac{2x}{3} - 16 = -6|
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Transform the equation’s operations into inverse operations. Make sure to also reverse the order in which the operations are usually completed. ||\begin{align} &x \to \times 2 \to \div 3 \to - 16\ = -6 \\ &x =\: \div 2 \leftarrow \times 3 \leftarrow + 16 \leftarrow -6 \end{align}||
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The operations are completed from right to left. ||\begin{align} x &= \div 2 \leftarrow \times 3 \leftarrow \color{red}{+ 16 \leftarrow -6}\\ x &= \div 2 \leftarrow \color{red}{\times 3 \leftarrow + 10}\\ x &= \color{red}{\div 2 \leftarrow 30} \\ x &= 15 \end{align}||
Conclusion: |x = 15.|
The hidden terms method, also called the cover up method, consists of hiding an algebraic term in order to find the value of the hidden term afterward.
The hidden terms method can be applied using the following approach:
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Cover up the part of the operator whose value is unknown.
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Repeat step 1, but for the part that was covered in this step.
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Repeat step 2, but for the part that was covered in this step.
Continue until the value of the variable is found.
Find the value of |x| in the following equation: |\displaystyle \frac{5x}{3} - 12 = 8.|
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Look for the value of |\displaystyle \frac{5x}{3}.|
Cover the term |\displaystyle \frac{5x}{3}| in the equation. ||\begin{align} \color{red}{?} - 12 &= 8\\ \color{red}{20} - 12 &= 8 \end{align}|| Observe that |\displaystyle \frac{5x}{3} = 20.|
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Next, look for the value of |5x.|
Hide the term |5x| in the equation. ||\begin{align} \frac{\color{red}{?}}{3} = 20\\ \frac{\color{red}{60}}{3} = 20 \end{align}|| Observe that |5x = 60.|
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Look for the value of |x.|
Hide the term |x| in the equation. ||\begin{align} 5 \times\ \color{red}{?}\ &= 60 \\ 5\times \color{red}{12} &= 60 \end{align}|| Observe that |x = 12.|
Conclusion: |x = 12.|
The trial-and-error method involves trying out different possible values for a variable and checking to see if they are solutions for the equation.
In the trial-and-error method, we randomly choose values for the variable and check to see if these values correspond to the solution of the equation. Although it’s simple to perform, this method has the disadvantage of being long and random. For that reason, it is better to master the other techniques to solve the equations more efficiently.
Find the value of |x| in the following equation: |\displaystyle \frac{x}{2} + 6 = 10.|
1st test: Replace |x| with |2.| ||\begin{align} \frac{\color{red}{2}}{2} + 6 &\overset{?}{=} 10 \\ 1 + 6 &\overset{?}{=} 10 \\ 7 &\neq 10 \end{align}|| The two sides of the equation are not equal, because |7 < 10.| Observe that the solution is greater than |2.|
2nd test: Replace |x| with |10.| ||\begin{align} \frac{\color{red}{10}}{2} + 6 &\overset{?}{=} 10 \\ 5 + 6 &\overset{?}{=} 10 \\ 11 &\neq 10 \end{align}|| The two sides of the equation are not equal to each other, because |11>10.| Observe that the solution is less than |10.|
3rd test: Replace |x| with |8.| ||\begin{align} \frac{\color{red}{8}}{2} + 6 &\overset{?}{=} 10 \\ 4 + 6 &\overset{?}{=} 10 \\ 10 &= 10 \end{align}||
The two sides of the equation are equal to each other. Conclusion: the solution is |x = 8.|
Verifying an equation’s solution is a process used to check the accuracy of the value of the variable found.
In order to verify a solution, simply replace the variable in the starting equation with the solution found.
The solution |x = 12| was obtained in the example on the cover-up method seen above. In order to check if this answer makes the initial equation true, simply replace the variable with the value of the solution.
||\begin{align} \frac{5x}{3} -12 &= 8 \\ \frac{5 \times \color{red}{12}}{3} -12 &\overset{?}{=} 8 \\ \frac{60}{3} -12 &\overset{?}{=} 8 \\ 20 -12 &\overset{?}{=} 8 \\ 8 &= 8 \end{align}||
Since both sides of the equation are equal, the conclusion that |x=12| is correct.
The various methods for solving equations presented above are general methods. To see specific methods for solving each type of function, consult the following links:
Pour valider ta compréhension à propos de la résolution de problèmes algébriques de façon interactive, consulte la MiniRécup suivante :
Pour valider ta compréhension à propos des mesures manquantes dans les figures planes de façon interactive, consulte plutôt la MiniRécup suivante :
Pour valider ta compréhension à propos des mesures manquantes dans les solides de façon interactive, consulte plutôt la MiniRécup suivante :