How to Solve Systems of Equations

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A system of equations refers to a set or collection of equations that share the same variables. The goal of solving systems of equations is to identify the location where the lines intersect when the equations are graphed. The $(x,y)$ ordered pair of this intersection point is considered the solution of the system. There are generally four methods for solving systems of equations: substitution, elimination, graphing, and matrices. The method that you select depends on the structure of the equations. For example, if the equations are already in slope-intercept form $(y=mx+b)$, then the graphing method is a convenient option.

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When solving systems of equations, you have a few options as far as which method you choose. The method that you select will depend on the structure of the equations. Selecting the strategy that is most convenient and efficient is of course the goal. The more you work with systems, the more trained your eye will become in selecting the best method. The four methods to choose from include substitution, elimination, graphing, and matrices. Each approach is quite different, but every strategy will ultimately arrive at the same goal, which is to locate the $(x,y)$ ordered pair where the lines intersect.

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Systems of equations have many useful real-world applications. For example, whenever you are given two unknown values, as well as information that connects the two unknown values, a system of equations can be set up in order to solve for the two unknowns. For example, let’s say the admission fee to an amusement park is $\$20$ for children and $\$30$ for adults. Let’s also say that $215$ people entered the park today, and that the total admission fees collected today was $\$\mathrm{3,500}$. We can solve for the number of children and adult tickets sold by setting up a system of equations. We have two unknown values, and we have enough information connecting the two values to set up our equations. To solve this problem, we first need to select a strategy (substitution, elimination, graphing, or matrices). Solving systems of equations is a skill that has many valuable applications. This is why the process of solving systems is considered a major branch of algebra.

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A linear equation can be graphed by plugging in values for $x$, and then calculating the corresponding $y$-values. These $x$– and $y$-coordinates can be graphed and will eventually form a line. When there is more than one equation, and they share the same variables, the equations are called a “system”.

An example of a system of equations:
$y=2x+5$
$3y=4x-8$

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The exciting thing about learning how to solve systems of equations is that the skill can be applied to almost any career! Anytime you are faced with a scenario where you have two unknown values, and enough information to compare the two values, setting up and solving a system of equations will allow you to identify the amount of each value. This is often useful in the field of finance, business, and sales. Specifically careers that involve calculations with costs, revenue, and profit.

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Solving systems of equations is a valuable skill to study because of its many applications in the real-world. Anytime we have two unknown values, and enough information to compare the two values, we can solve for the unknowns by setting up a system of equations. Remember, the solution of a system of equations is the value for each variable that makes both of the equations true. This skill has applications in many areas of our daily lives. For example, we can use a system of equations to determine how many calories we burn using different machines at the gym. If we use the rowing machine for $30$ minutes and the treadmill for $20$ minutes, and we burn a total of $430$ calories, this can be set up as equation #1: $30r+20w=430$. If we go to the gym the following day and use the rowing machine for $50$ minutes and the treadmill for $10$ minutes, burning a total of $600$ calories, this can be set up as equation #2: $50r+10w=600$. We have two unknown values (how many calories we burn per minute on each machine), and we have enough information comparing both of the values. This means that we can solve the problem by setting up and solving a system of equations.

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Solving systems of equations by substitution follows three basic steps.

Step 1: Solve one equation for one of the variables.
Step 2: Substitute this expression into the other equation, and solve for the missing variable.
Step 3: Substitute this answer into one of the equations in order to solve for the other variable.

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The following system of equations can be solved in three steps.

$2x+y=1$
$x-2y=8$

Step 1: Solve one equation for one of the variables.

Look for the equation with a coefficient of $1$ or $-1$. This makes the solving process quicker. $x-2y=8$ has a coefficient of $1$ for $x$, so let’s solve for $x$. When $2y$ is added to both sides of the equation $x-2y=8$ becomes $x=8+2y$.

Step 2: Substitute this expression into the other equation, and solve for the missing variable.

$x=8+2y$ has been solved for $x$, so plug this expression into the other equation for $x$. $2x+y=1$ becomes $2(8+2y)+y=1$.

Now we have an equation with only the y-variable. At this point we can solve for y. First we distribute, so $2(8+2y)+y=1$ becomes $16+4y+y=1$. Now we combine like-terms. $16+4y+y=1$becomes $16+5y=1$. Now subtract $16$ from both sides, divide by $5$, and we see that $y=-3$.

Step 3: Substitute this answer into one of the original equations in order to solve for the other variable.

Now that we know$y=-3$, we can solve for $x$. Plug $-3$ in for $y$, in either equation. Let’s use the first equation $2x+y=1$. When $-3$ is plugged in for $y$, the equation becomes $2x+(-3)=1$. When we solve for $x$, we see that $x=2$.

We know that $x=2$ and $y=-3$. This means that the ordered pair $(2,-3)$ is the solution of the system of equations. This location is where the lines will cross each other when graphed.

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Systems of equations can be solved using elimination. This method follows three basic steps. Step 1: Manipulate the equations so that one variable will cancel out when the equations are added or subtracted. Step 2: Add or subtract the equations so that one variable is eliminated. Solve for the remaining variable. Step 3: Plug the solved variable back into one of the original equations in order to solve for the other variable.

For example, the following system can be solved by elimination.
$3x-4y=-5$
$5x+8y=-1$

Step 1: Manipulate the equations so that one variable will cancel out when the equations are added or subtracted.
In this case, we can multiply the first equation by $2$, so that the $y$-variables will cancel out when the equations are added. $3x-4y=-5$ becomes $6x-8y=-10$.

Step 2: Add or subtract the equations so that one variable is eliminated. Solve for the remaining variable.
$6x-8y=-10$
$5x+8y=-1$
When the equations are added, the result is $11x=-11$. Now we can solve for $x$ by dividing both sides of the equation by $11$. $x=-1$.

Step 3: Plug the solved variable back into one of the original equations in order to solve for the other variable.Let’s plug $-1$ in for $x$ in the first equation $3x-4y=-5$. The equation now becomes $3(-1)-4y=-5$ which simplifies to$y=\frac{1}{2}$.

$x=-1$ and $y=\frac{1}{2}$ which means that the solution to the system of equations is the ordered pair $(-1,\frac{1}{2})$. If the two original equations were graphed, this ordered pair is where the two lines would intersect.

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Sometimes a system will have more than two equations. For example the following system has three equations.
$x+2y-z=2$
$2x-3y+z=-1$
$5x-y-2z=-3$

The process for solving a system of three equations will be the same as the process for two equations. However, our first objective is to go from three equations down to two. This can be achieved by adding or subtracting two of the equations. Let’s add the first two equations.

The variable $z$ has dropped out, and now we only have two equations which means we are on the right track. However, we need to incorporate the third equation into the mix. Let’s do this by multiplying the second equation by $2$, and then adding this to the third equation.

The second equation becomes $4x-6y+2z=-2$ and when we add this to the third equation, we get $9x-7y=-5$ (The $z$s dropped out!).

At this point, we are simply solving a two equation system with two variables.

$3x-y=1$ (Multiply this equation by $-7$ so that the $y$-variables will be eliminated when the equations are added together)

Now plug in $1$ for $x$ in any of the equations.
$3(1)-y=1$
$y=2$

Finally, plug in $x=1$ and $y=2$ to any of the equations and solve for $z$. If these values are plugged into the first equation, $x+2y-z=2$ becomes $(1)+2(2)-z=2$, which simplifies to $z=3$.

Remember, when you have a system with three equations and three variables, the big idea is to work your way down to two equations with two variables, and eventually one equation with one variable. The more you practice these elimination problems the easier they will become!

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When solving a system of equations using the elimination method, the equations can be added or subtracted in order to eliminate a variable. For example, if you notice that both equations have the same variable with the same sign, then elimination using subtraction is likely the most efficient method.

For example, see if you can identify why subtraction with elimination is the best option for solving the following system.

You’re correct if you noticed that both equations have the same variable with the same sign: $-3y$ and $-3y$. This means that if one of the equations is subtracted from the other, the term will drop out, leaving only one variable (that’s the goal).

Now $-3$ can be plugged into one of the original equations in order to solve for $y$.

$2x-3y=9$ becomes $2(-3)-3y=9$ which simplifies to $y=-5$. $(-3,-5)$ is the solution.