Methods of solving system of linear equations

Another way of solving a linear system is to use the elimination method. In the elimination method you either add or subtract the equations to get an equation in one variable.

When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.

Example

\begin{cases} 3y+2x=6\\ 5y-2x=10 \end{cases}

We can eliminate the \(x\)-variable by addition of the two equations.

\begin{cases} 3y+2x=6 \\ \underline{+\: 5y-2x=10} \end{cases} 

$$=8y\: \: \: \: \; \; \; \; =16$$

$$\begin{matrix} \: \: \: y\: \: \: \: \: \; \; \; \; \; =2 \end{matrix}$$

The value of \(y\) can now be substituted into either of the original equations to find the value of \(x\)

$$3y+2x=6$$

$$3\cdot {\color{green} 2}+2x=6$$

$$6+2x=6$$

$$x=0$$

The solution of the linear system is \((0, 2)\).

To avoid errors make sure that all like terms and equal signs are in the same columns before beginning the elimination.

If you don't have equations where you can eliminate a variable by addition or subtraction you directly you can begin by multiplying one or both of the equations with a constant to obtain an equivalent linear system where you can eliminate one of the variables by addition or subtraction.

Example

\begin{cases} 3x+y=9\\ 5x+4y=22 \end{cases}

Begin by multiplying the first equation by \(-4\) so that the coefficients of \(y\) are opposites

\begin{cases} \color{green}{-4} \cdot (3x + y) = \color{green}{-4} \cdot 9\\ 5x + 4y = 22 \end{cases}

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Systems of equations are multiple equations that all have a common solution. Students encounter these systems of equations when there are multiple ‘unknowns’ - or variables - that have not been given to them yet. When this happens, the goal for students is to use given information in the equations to solve for all variables. 

To solve systems of equations, it is helpful for students to have a background understanding of simple algebraic equations, variables, and graphing linear equations.

How do I solve a system of equations?

There are three methods used to solve systems of equations: graphing, substitution, and elimination. 

To solve a system by graphing, you simply graph the given equations and find the point(s) where they all intersect. The coordinate of this point will give you the values of the variables that you are solving for. This is most efficient when the equations are already written in slope-intercept form.

The next method is substitution. Substitution is best used when one of the equations is in terms of one of the variables such as y=2x+4, but the equations can always be manipulated. The first step in this method is to solve one of the equations for one variable. Once an expression for the variable is found, substitute or plug in the expression into the other equation where the original variable was to solve for the number value of the next variable. The final step is to substitute the number value that was found in for its corresponding variable in the original equation. 

The third method is elimination. Elimination is adding the equations together in order to create an equation with only one variable. This can only be done when the coefficients of one variable in both equations are opposites and will cancel each other out once added together. Elimination is best used when this is already occurring in the equations, but the equations can also be manipulated into creating common coefficients by either multiplying or dividing equations by a certain number. The next step would be to use the equation that we created to find the value of the variable and then plug that value back into an original equation to find the remaining variable. 

Here’s an example of a problem where solving a system of equations is necessary:

Logan has answered 0.8 as many math questions as Spanish questions, and he’s answered 5 more English questions than Spanish questions. If Logan has answered 33 questions in total, how many math questions has Logan answered?

How do I solve this problem?

The first step is to create equations from the word problem. To do this, we must assign variables to each unknown part of the problem. The variables x, y, and z will represent the amount of math, Spanish, and English questions that Logan has answered respectively.

As Logan has answered 0.8 times as many math questions as Spanish ones, the equation to represent this would be 0.8y=x. The second equation would be z=y+5 to represent how Logan has answered five more English questions than Spanish ones. The final equation would be x+y+z=33 to represent how Logan answered 33 questions in total.

Looking back at the original question, the goal of this problem is to find how many math questions Logan has answered. Since the first equation we found was 0.8y=x, we can see that we only need the y variable to find the value of x, or the amount of math questions that were answered. Because there are two equations already solved in terms of two variables, 0.8y=x and z=y+5, substitution would be the most efficient method. To do this method, we would substitute these equations for the x and y variables of the third equation which would give us (0.8y)+y+(y+5)=33.

The next step would be to solve this equation for the y variable by combining like terms: 2.8y=28 which would give us y=10 or 10 Spanish questions answered. Now that we have found the value of the y variable, we can plug it back into the equation, 0.8y=x, to find the value of x. By substituting y with its value of 10, we would get 0.8(10)=x which would give us a value of 8 for x.

What’s the answer?

Logan answered 8 math questions.

What concepts did we use?

To solve that example problem, we used several different math concepts. The first one we used was how to write equations from word problems. Through our understanding of the problem, we were able to assign each unknown aspect of the problem a variable and then create equations based off of their relationships in the problem which we would then recognize as a system of equations.

The second concept that we used is solving the system. The equations weren’t written in slope-intercept form, so graphing would not have been an efficient method. There were no variables that were opposite each other, so we also ruled out the elimination method. By recognizing substitution as the best method to use, we were able to efficiently use our math skills to solve for the unknown variables in a system of equations.

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