How to Solve Systems Using Linear Combinations

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Last Updated: February 19, 2024 References

A “system of equations” is type of math problem in which you have two or more separate equations and you need to find the values of two or more variables. In general, to be able to find a solution, you need to have as many different equations as the number of variables that you wish to find. (There are advanced problems where the number of equations and the number of variables do not match, but that will not be addressed here.)

Method 1

Method 1 of 5:

Arranging the Equations to Begin Solving

1
Recognize the standard format. In algebra, the “standard format” for an equation is written as $Ax+By=C$ $Ax+By=C$ .^{[1]
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Research source} When written in this format, the letters A, B and C are commonly chosen to represent numerical values, while x and y are the variables that you need to solve.
- You could easily work with different variables, but the structure of the standard format will be the same. For example, if you are solving a business-related problem about selling hats and scarves to calculate the total number of items sold, you might choose the variable $h$ to represent the number of hats and $s$ to represent the number of scarves. Your standard format in this case would look like $Ah+Bs=T$ . The steps for solving the problem will still be the same.
2
Rearrange your equations to put them into standard format. This may require you to combine similar terms, if each variable appears in the equation more than once, for example.^{[2]
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Research source} You will also need to move the terms so they appear in the proper order.^{[3]
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- For example, given the equation $2x+y+2y+3x+1=4$ $2x+y+2y+3x+1=4$ , you need to perform the following steps to get to standard format:
  - $2x+y+2y+3x+1=4$ (given equation)
  - $5x+3y+1=4$ (combine like terms)
  - $5x+3y=3$ (subtract 1 from both sides)
- You may be familiar with seeing linear equations in the form $y=mx+b$ $y=mx+b$ . This is called the “slope-intercept” form of a line. It is useful for different purposes. It could be used to solve the system by linear combinations, but the standard format Ax+By=C is preferred. If you have your data in the slope-intercept form, you will need to rewrite it algebraically into standard format as follows:
  - $y=mx+b$ (given slope-intercept form)
  - $y-mx=b$ (subtract mx from both sides)
  - - $mx+y=b$ (rearrange terms to get x first)
  - A=-m, B=1, C=b (redefine terms for standard format)
3
Write your equations so the variables line up. It is helpful to write your equations with one directly over the other, so the similar terms line up.
- For example, if you have the two equations, in standard format, of $2x-2y=5$ $2x-2y=5$ and $3x+2y=8$ $3x+2y=8$ , write them in two rows as:
  - $2x-2y=5$
  - $3x+2y=8$
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Method 2

Method 2 of 5:

Using Linear Combinations if a Pair of Coefficients Match

1
Examine the equations in standard format. When you have your equations written in standard format, lined up so the similar terms are aligned, check the coefficients. You are looking for one pair of coefficients that match.^{[4]
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- For example, consider these two equations:
  - $2x+y=5$
  - $2x-3y=1$
- You should see very quickly that the term $2x$ appears identically in each equation.
- Be very careful when matching the terms. Look for the signs (plus or minus) to match as well. For this method of solving, the terms $2x$ and $-2x$ are NOT considered to be the same.
- If your system does not have a matching pair of coefficients, you cannot use this method for solving. You will need to go on to the next method.
2
Subtract corresponding terms. Working across the system from left to right, subtract each term of the second equation from the corresponding term of the first equation.
- It may be helpful simply to draw a long horizontal line across the bottom of the two equations and subtract downward, as you would with any ordinary subtraction problem.
  - $2x+y=5$
  - $2x-3y=1$
  - ------------------------
  - $0x+4y=4$
3
Write out the result. If one of your terms matched exactly, as it should, and you subtracted correctly, then one of the variables should be eliminated from the problem. Rewrite what you have left as a single equation.
- In the example above, you should be left with $4y=4$ .
- Because one of the variables gets eliminated in this method, some textbooks will refer to this as the “elimination” method of solving a system of equations.
4
Solve for the remaining variable. What you have left should be a fairly simple, one-variable equation. Solve it by dividing both sides of the equation by the coefficient.^{[5]
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- In the example above, divide both sides of $4y=4$ by 4. You will be left with the solution $y=1$ .
5
Replace that solution into one of your original equations. Take that solution, in our example y=1, and substitute it in place of $y$ $y$ in either one of the original equations.
- In this case, we can choose the first example, $2x+y=5$ . When you replace the variable with its solution, you will have $2x+1=5$ .
6
Solve for the remaining variable. Use basic algebraic steps to solve for the remaining variable. Remember that whatever action you do to one side of the equation, you must also do to the other side.^{[6]
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Research source} For example:
- $2x+1=5$ (original equation)
- $2x=4$ (subtract 1 from both sides)
- $x=2$ (divide both sides by 2 to get solution)
7
Check your two solutions. Verify that you have done the work correctly by checking your solutions. You should be able to place your two solutions, in this example $x=2$ $x=2$ and $y=1$ $y=1$ , into each of the original equations. When you then simplify the equations, you will get true statements.
- For example, check the first equation as follows:
  - $2x+y=5$ (original equation)
  - $2*2+1=5$ (insert values for x and y)
  - $4+1=5$ (simplify multiplication)
  - $5=5$ (simplify addition, to get solution)
  - The true statement 5=5 shows the solution is correct.
- Check the second equation as follows:
  - $2x-3y=1$ (original equation)
  - $2*2-3*1=1$ (insert values for x and y)
  - $4-3=1$ (simplify multiplication)
  - $1=1$ (simplify subtraction, to get solution)
  - The true statement 1=1 shows the solution is correct.
8
Write out your solution. The final solution, which you have proven to work in both equations, is $x=2$ $x=2$ and $y=1$ $y=1$ .^{[7]
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- If you are working on graphing linear functions, you may also write your solution as an ordered pair. Thus, for this example, you would write $x=2$ and $y=1$ in the form $(2,1)$ .
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Method 3

Method 3 of 5:

Using Linear Combinations if a Pair of Coefficients are Opposites

1
Examine the equations in standard format. Set up your two equations in standard format and look at the coefficients of each of your variables. You are looking for the circumstance where the numbers are the same but the signs are different.^{[8]
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Research source}
- Consider this example:
  - $x-3y=5$
  - $2x+3y=19$
- By examination, you should see that the first equation contains the term $-3y$ , while the second equation contains the term $3y$ . These two terms are opposites of each other.
2
Add corresponding terms. Working across the system from left to right, add each term of the first equation to the corresponding term of the second equation. It may be helpful simply to draw a long horizontal line across the bottom of the two equations and add downward, as you would with any ordinary addition problem.
- The above example works out as follows:
  - $x-3y=5$
  - $2x+3y=19$
  - -------------------------
  - $3x=24$
3
Write out the result. Because you were adding, and one of your terms contained opposites, then one of the variables should be eliminated from the problem. Rewrite what you have left as a single equation.
- In the example above, the $y$ variable was eliminated. The remaining equation is $3x=24$ .
- Because one of the variables gets eliminated in this method, as with the prior subtraction method, some textbooks will refer to this as the “elimination” method of solving a system of equations.
4
Solve for the remaining variable. What you have left should be a fairly simple, one-variable equation. Solve it by dividing both sides of the equation by the coefficient.
- In the example above, divide both sides of $3x=24$ by 3. You will be left with the solution $x=8$ .
5
Solve the second variable. Take that solution, in our example x=8, and substitute it in place of $x$ $x$ in either one of the original equations.
- Choose the first equation:
  - $x-3y=5$ (original equation)
  - $8-3y=5$ (insert value of x)
  - - $3y=$ - $3$ <subtract 8 from both sides)
  - $y=1$ (divide both sides by -3, to get solution)
6
Check your two solutions. Verify that you have done the work correctly by checking your solutions. You should be able to place your two solutions, in this example $x=8$ $x=8$ and $y=1$ $y=1$ , into each of the original equations. When you then simplify the equations, you will get true statements.
- For example, start with the first equation:
  - $x-3y=5$ (original equation)
  - $8-3*1=5$ (insert values of x and y)
  - $8-3=5$ (simplify multiplication)
  - $5=5$ (simplify subtraction to get solution)
  - The true statement 5=5 shows that the solution is correct.
- Now try the second equation:
  - $2x+3y=19$ (original equation)
  - $2*8+3*1=19$ (insert values of x and y)
  - $16+3=19$ (simplify multiplication)
  - $19=19$ (simplify addition to get solution)
  - The true statement 19=19 shows that the solution is correct.
7
Write out your solution. The final solution, which you have proven to work in both equations, is $x=8$ $x=8$ and $y=1$ $y=1$ .^{[9]
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Research source}
- If you are working on graphing linear functions, you may also write your solution as an ordered pair. This for this example, you would write $x=8$ and $y=1$ in the form $(8,1)$ .
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Method 4

Method 4 of 5:

Using Linear Combinations for any Coefficients

1
Examine the equations in standard format. It is more likely that your system of equations will not have a pair of matching or opposite coefficients. When you line up the two equations and compare coefficients, unless two coefficients (the A and B of the standard format) match exactly, you need to take a couple extra steps.^{[10]
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Research source}
- For example, consider these two initial equations:
  - $3x+2y=6$
  - $8x-4y=2$
- When you examine them, there are no matching coefficients for similar terms. That is, the 3x does not match the 8x, and the 2y does not match the -4y. There is also no pair of opposites.
2
Create a pair of matching or opposite coefficients. Examine the two equations and decide what number you could use to multiply one of the equations, to create a pair of matching or opposite coefficients. For example, given the system $3x+2y=6$ $3x+2y=6$ and $8x-4y=2$ $8x-4y=2$ , you should be able to see that the first equation contains a term $2y$ $2y$ and the second equation contains a term - $4y$ $4y$ . If you double the first term, you will have a pair of opposite coefficients.
- Multiply each term of the equation to create a new equation for solving. In this example, multiply each term of the first equation by $2$ . This will turn the original equation $3x+2y=6$ into $6x+4y=12$ . Notice that you now have a pair of opposite coefficients in the $y$ terms of $4y$ and - $4y$ .
- In some cases, you may need to do a double multiplication, or use a fraction. For example, in the system $2x-3y=2$ $2x-3y=2$ and $5x+2y=1$ $5x+2y=1$ , there are no coefficients that are simple integer multiples of each other. You could multiply the first equation by $5/2$ $5/2$ to create $5x-{\frac {15}{2}}y=5$ $5x-{\frac {15}{2}}y=5$ , and now the $x$ $x$ coefficients are ready to be canceled. Alternatively, if you prefer not to work with fractions, you could multiply the first equation by 5 and the second equation by 2. This would create two completely new equations, as follows:
  - $2x-3y=2$ (first original equation)
  - $5x+2y=1$ (second original equation)
  - Now multiply the first equation by 5, and the second equation by 2
  - $5*(2x-3y=2)$ →→ $10x-15y=10$
  - $2*(5x+2y=1)$ →→ $10x+4y=2$
3
Either add or subtract the two new equations. If you have created a matching pair of coefficients, you will subtract terms to eliminate one variable. If you have created a pair of opposite coefficients, you will add terms to eliminate one variable. Consider the following example:
- - $6x+4y=12$ (first equation)
  - $8x-4y=2$ (second equation)
  - ----------------------
  - $14x=14$ (add two equations together to cancel y terms)
  - $x=1$ (divide by 14 to get solution)
4
Replace that solution into one of your original equations. Take that solution, in our example x=1, and substitute it in place of $x$ $x$ in either one of the original equations. This works as follows:
- $3x+2y=6$ (original equation)
- $3*1+2y=6$ (insert x value)
- $3+2y=6$ (simplify multiplication)
- $2y=3$ (subtract 3 from both sides)
- $y={\frac {3}{2}}$ (divide both sides by 2)
5
Check your two solutions. Verify that you have done the work correctly by checking your solutions. You should be able to place your two solutions, in this example $x=1$ $x=1$ and $y={\frac {3}{2}}$ $y={\frac {3}{2}}$ , into each of the original equations. When you then simplify the equations, you should get true statements.
- For example, check the first equation:
  - $3x+2y=6$ (original equation)
  - $3*1+2*{\frac {3}{2}}=6$ (insert x and y values)
  - $3+3=6$ (simplify multiplication)
  - $6=6$ (simplify addition to get solution)
  - The true statement $6=6$ shows that the solution is correct.
- Now check the second equation, as follows:
  - $8x-4y=2$ (original equation)
  - $8*1-4*{\frac {3}{2}}=2$ (insert x and y values)
  - $8-6=2$ (simplify multiplication)
  - $2=2$ (simplify subtraction)
  - The true statement $2=2$ shows that the solution is correct.
6
Write out your solution. The final solution, which you have proven to work in both equations, is $x=1$ $x=1$ and $y={\frac {3}{2}}$ $y={\frac {3}{2}}$ .^{[11]
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Research source}
- If you are working on graphing linear functions, you may also write your solution as an ordered pair. This for this example, you would write $x=1$ and $y={\frac {3}{2}}$ in the form $(1,{\frac {3}{2}})$ .
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Method 5

Method 5 of 5:

Dealing with Special Circumstances

1
Recognize identical equations as having infinite solutions.^{[12]
X
Research source} In some circumstances, your system of linear equations may have infinite solutions. This means that any pair of values that you insert into the two variables will make the two equations correct. This happens when the two equations are really just algebraic variations of the same, single equation.
- For example, consider these two equations:
  - $2x+8y=18$
  - $x+4y=9$
- If you begin working on this system and try creating a pair of matching coefficients, you will find that by multiplying the second equation by 2 you will create the equation $2x+8y=18$ . This is an exact match of the first equation. If you proceed through the steps, you will eventually get the result $0=0$ .
- A solution of 0=0 means that you have “infinite” solutions or you can simply say that the two equations are identical.
- If you consider this system graphically and plot the lines that are represented by the two equations, the “infinite” solution means that the two lines lie exactly one on top of the other. It is really only one line.
2
Find systems with no solution.^{[13]
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Research source} Occasionally you may have a system in which the two equations, when written in standard form, are nearly identical except that the constant term C is different. Such a system has no solution.
- Consider these equations:
  - $4x+2y=6$
  - $2x+y=4$
- At first glance, these look like very different equations. However, when you begin solving and multiply each term of the second equation by 2 to try to create matching coefficients, you will wind up with the two equations:
  - $4x+2y=6$
  - $4x+2y=8$
- This is an impossible situation, since the expression $4x+2y$ cannot equal both 6 and 8 at the same time. If you were to try solving this by subtracting the terms, you would reach the result $0=-2$ , which is an incorrect statement. In such a circumstance, your response is that there is no solution to this system.
- If you consider what this system means graphically, these are two parallel lines. They will never intersect, so there is no single solution to the system.
3
Use a matrix for systems with more than two variables.^{[14]
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Research source} It is possible for a system of linear equations to have more than two variables. You may have 3, 4, or as many variables as the problem dictates. Finding a solution to the system means finding a single value for each variable that makes each equation in the system correct. To find a single, unique solution, you must have as many equations as you have variables. Thus, if you have the variables $x,y$ $x,y$ and $z$ $z$ , you need three equations.
- Solving a system of three or more variables can be done using the linear combinations explained here, but that gets very complicated. The preferred method is using matrices, which is too advanced for this article. You may wish to read Use a Graphing Calculator to Solve a System of Equations.
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Given the two equations 6x+ 2y = 2 and 8x + 3y = 14, explain how knowing how to find the least common multiple (LCM) of two numbers can help you solve the system of equations presented here by eliminating the x-terms.

Community Answer

These equations are already written in standard format, Ax+By=C. However, the x coefficients do not match, so they are not ready to be eliminated. To eliminate the x terms, you need to do a multiplication step. The quickest way to multiply would be to multiply the first equation by the x-coefficient from the second equation, and multiply the second equation by the x-coefficient from the first equation. This would give you the equations 48x+16y=16 and 48x+18y=84. These are very large, bulky numbers. If you recognize that 6 and 8 have a least common multiple of 24, you could choose different numbers to multiply. Multiply the first equation by 4 and the second equation by 3, and you will create 24x+8y=8 and 24x+9y=42. You can now solve the system more easily.

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How do i solve 2×+5y=1/3×-2y=8

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This is a strange way of writing the problem. By using two equals signs in a line like this, you have combined two different equations. Breaking these apart gives 2x+5y=8 and 1/3 x -2y=8. Because each equation equals 8, you can just drop the 8 and put the two equations as equal to each other, so you have 2x+5y=1/3 x-2y. Because you have two variables here, and only one equation, you cannot get a single solution. The best you can do is simplify the equation by combining like terms. This will give 1 2/3 x = -7y. If you wish to write this in a slope-intercept form to describe a line, you could divide both sides by 7, and switch sides, to get y= -5/21 x. My guess is that you probably copied your problem incorrectly. Check it again.

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This is just the elimination method. Is there any other simpler method?

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There's also the substitution method, but it is only simpler in some cases.

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Tips

Note that linear combinations is only one way to solve a system of equations. You can also solve a system by substitution. For help on that, see Solve Systems of Algebraic Equations Containing Two Variables.

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Warnings

Always pay careful attention to the signs of the coefficients. Common mistakes occur when people miss negative signs. Remember that if the signs match, then you will subtract terms to eliminate a variable. If the signs are opposites, then you will add to eliminate a variable.

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