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A system of equations is a set of two or more equations. Systems can be solved graphically and algebraically. In this section, one algebraic method will be explored.

Systems of linear equations can be solved algebraically by elimination. The goal of this method is to eliminate one of the variables, resulting in a one-variable equation. This is done by intentionally combining the two equations. For example, the following system can be solved in this way. ${2a+b=4b=3a−1 $

Analyze coefficients

In most cases, one variable is easier to eliminate than the other. This depends on the coefficients. Therefore, the coefficients should be analyzed to find which variable to eliminate. Variables that have the same coefficient, disregarding the sign, in both equations are the easiest to eliminate. Next is whichever variable requires the least amount of manipulation to get the same coefficient in both equations. In
${2a+b=4b=3a−1, $
the variable $b$ has the coefficient $1$ in both equations. Therefore, it is the preferred variable to eliminate.

Manipulate equations

Before the chosen variable can be eliminated, the equations most likely have to be manipulated in some way. The goal here is to make sure that the coefficients of the chosen variable are the same, ignoring the sign. It is also a good idea to move all variables to one side, and all constants to the other. For the example, the term $3a$ in the second equation should be moved to the left-hand side to resemble the first equation.

Combine equations

The equations can now be combined, by either adding or subtracting one of the equations from the other, whichever will lead to eliminating one of the variables. In the example,
${2a+b=4-3a+b=-1, $
the coefficients of $b$ are the **same sign**. Therefore, **subtracting** one equation from the other will eliminate $b.$ The second equation will now be subtracted from the first, though it is an arbitrary choice. This is done by subtracting the LHS of the second equation from the first, and doing the same for the RHS.

$− 2a+b-3a+b5a+0 === 4-15 $

Solve the resulting equation

The resultant equation can be solved using inverse operations. This gives the value of one of the two variables in the system. In the example, the resulting equation can be solved by dividing by $5$ on both sides.
$5a=5⇔a=1$
Thus, $a=1.$

Substitute variable value into either original equation to solve for the other variable

The value of the variable just found can now be substituted into either of the two original equations to create yet another one-variable equation. Solving this new equation gives the value of the second variable, thereby solving the system completely. For the example, $a=1$ will be substituted into the second equation,
$b=3a−1.$
The second equation is chosen since it will require fewer steps to solve than the first equation. However, it is an arbitrary choice.

The value of both variables have now been found, $a=1,$ and $b=2.$ Thus, the solution to the system is $(1,2).$

Solve the following system using elimination. ${-2x+y=49x−3y=-6 $

Show Solution

When solving a system of equations by elimination, we should first analyze the coefficients to decide which variable to eliminate. One of the variables must have the same coefficient in both equations, disregarding their signs. In the system ${-2x+y=49x−3y=-6, $ there is no such variable. Therefore, we now have to manipulate the equations so that this condition is met. This is done by multiplying either one or both equations by some number. Notice that if we multiply the first equation by $3,$ the coefficients of $y$ will be $3$ and $-3.$ Therefore, we'll choose to eliminate the variable $y.$

Since the coefficients of $y$ have different signs, the variable will be eliminated by adding the two equations.

$+ -6x+3y-9x−3y3x+-0 === 12-6-6 $

The resulting equation is $3x=6.$ Solving this equation gives the $x$-coordinate of the solution to the system. $3x=6⇔x=2$ To find the corresponding $y$-coordinate, we substitute $x=2$ into either original equation. For simplicity, let's choose $-2x+y=4.$ The solution to the system is the point $(2,8).$ {{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

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