The comparison method is a method that algebraically solves a system when the two equations are of the form |y=ax+b|.
Solving a system of equations by comparison is generally preferred when the same variable in both equations is already isolated. In other words, use this method when the system is of the following form. ||\begin{cases}y = a_1x + b_1 \\ y = a_2x + b_2 \end{cases}||
It is still possible to use this method if the two dependent variables are not isolated in their respective equations. Just carry out some algebraic manipulations to isolate these variables before proceeding to compare the equations.
Solving a system of equations consists of finding the value of |x| when the value of |y| is the same in both equations. Since |y=y|, the transitive property of the equals sign gives us the following single-variable equation: |a_1x+b_1=a_2x+b_2|. This is where we compare the two sides of the equation.
It does not have to be the dependent variable (usually, |y|) that is isolated in the two equations. It just has to be the same variable that is isolated, regardless if it is the dependent variable or the independent variable.
To solve of a system of equations using the comparison method, follow these steps.
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For a word problem, define the variables and translate the problem into a system of equations.
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Isolate the same variable in both equations, if necessary.
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Make a single-variable equation by comparing the two algebraic expressions.
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Solve the equation.
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Substitute the value found in step 4 into one of the initial equations to find the value of the second variable.
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Verify the answer by substituting the values obtained for both variables into each of the initial equations.
Consider the following system of linear equations.
||\begin{cases}y + 4 = 3x - 1 \\ y=2x+2 \end{cases}||
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For a word problem, define the variables and translate the problem into a system of equations
As the system is already given, this step is not necessary. -
Isolate the same variable in both equations, if necessary
In this case, it is easier to isolate the |y| in the first equation only since it is already isolated in the second.
||\begin{align}y + 4 &= 3x - 1\\
y + 4 \color{red}{- 4} &= 3x - 1 \color{red}{- 4}\\
y &= 3x-5\end{align}|| -
Make a single-variable equation by comparing the two algebraic expressions
||\begin{align}y &= y \\ 3x - 5 &= 2x + 2\end{align}|| -
Solve the equation
Isolate |x| to determine its value.
||\begin{align}3x - 5 \color{red}{+ 5} &= 2x + 2 \color{red}{+ 5}\\
3x &= 2x + 7\\
3x \color{red}{- 2x} &= 2x + 7 \color{red}{- 2x}\\
x& = 7\end{align}|| -
Substitute the value found in step 4 into one of the initial equations to find the value of the second variable
Take |y=2x+2.|
||\begin{align}y& = 2x +2\\
y& = 2(7) +2\\
y &= 14 +2\\
y &= 16\end{align}|| -
Verify the answer by substituting the values obtained for the variables into each of the initial equations
||\begin{align} y &= 2x + 2 & y+4 &= 3x-1\\
(16)&=2(7)+2 & (16)+4 &= 3(7)-1\\
16&=14+2 & 20&=21-1\\
16&=16 & 20&=20 \end{align}||
Since the values make the two equations true, the solution to the system of equations is the point |(7, 16).|
Consider the following system of linear equations.
||\begin{cases}x = -5y + 9 \\ x=y+3 \end{cases}||
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For a word problem, define the variables and translate the problem into a system of equations
As the system is already given, this step is not necessary. -
Isolate the same variable in both equations, if necessary
Since the variable |x| is isolated in both equations, move on to the next step. -
Make a single-variable equation by comparing the two algebraic expressions ||\begin{align}x &= x \\ -5y+9 &= y+3\end{align}||
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Solve this equation
Isolate |y| to determine its value. ||\begin{align}-5y+9 \color{red}{-9} &= y+3 \color{red}{-9}\\
-5y &= y-6\\
-5y \color{red}{- y} &= y-6 \color{red}{- y}\\
-6y& = -6\\
-6y\color{red}{\div -6}&=-6\color{red}{\div -6}\\
y&=1\end{align}|| -
Substitute the value found in step 4 into one of the initial equations to find the value of the second variable
Take |x=-5y+9|. ||\begin{align}x& =-5y +9\\
x& = -5(1) +9\\
x&= -5+9\\
x& = 4\end{align}|| -
Verify the result by substituting the values obtained for the variables into each of the initial equations ||\begin{align} x &= -5y+9 & x &= y+3\\
(4)&=-5(1)+9 & (4)&= (1)+3\\
4&=-5+9 & 4&=4\\
4&=4 \end{align}||
Since the values make the two equations true, the solution to the system of equations is the point |(4,1)|.
Pour valider ta compréhension à propos de la résolution de systèmes d'équations à l'aide de la méthode de comparaison, de réduction ou de substitution de façon interactive, consulte la MiniRécup suivante.
