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How to Solve for Both X & Y

How to Add Miles Together

Solving for two variables (normally denoted as "x" and "y") requires two sets of equations. Assuming you have two equations, the best way for solving for both variables is to use the substitution method, which involves solving for one variable as far as possible, then plugging it back in to the other equation. Knowing how to solve a system of equations with two variables is important for several areas, including trying to find the coordinate for points on a graph.

Write out the two equations that have the two variables you want to solve. For this example, we will find the value for "x" and "y" in the two equations "3x + y = 2" and "x + 5y = 20"

Solve for one of the variables in on one of the equations. For this example, let's solve for "y" in the first equation. Subtract 3x from each side to get "y = 2 - 3x"

Plug in the y value found from the first equation in to the second equation in order to find the x value. In the previous example, this means the second equation becomes "x + 5(2- 3x) = 20"

Solve for x . The example equation becomes "x + 10 - 15x = 20," which is then "-14 x + 10 = 20." Subtract 10 from each side, divide by 14 and you have end up with x = -10/14, which simplifies to x = -5/7.

Plug in the x value in to the first equation to find out the y value. y = 2 - 3(-5/7) becomes 2 + 15/7, which is 29/7.

Check your work by plugging in the x and y values in to both of the equations.

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Drew Lichtenstein started writing in 2008. His articles have appeared in the collegiate newspaper "The Red and Black." He holds a Master of Arts in comparative literature from the University of Georgia.

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How to Solve Systems of Algebraic Equations Containing Two Variables

Last Updated: February 10, 2023 References

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 61 people, some anonymous, worked to edit and improve it over time. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been viewed 958,388 times. Learn more...

In a "system of equations," you are asked to solve two or more equations at the same time. When these have two different variables in them, such as x and y, or a and b, it can be tricky at first glance to see how to solve them. [1] X Research source Fortunately, once you know what to do, all you need is basic algebra skills (and sometimes some knowledge of fractions) to solve the problem. If you are a visual learner or if your teacher requires it, learn how to graph the equations as well. Graphing can be useful to "see what's going on" or to check your work, but it can be slower than the other methods, and doesn't work well for all systems of equations.

Using the Substitution Method

• This method often uses fractions later on. You can try the elimination method below instead if you don't like fractions.

• 4x = 8 - 2y
• (4x)/4 = (8/4) - (2y/4)

• You know that x = 2 - ½y .
• Your second equation, that you haven't yet altered, is 5x + 3y = 9 .
• In the second equation, replace x with "2 - ½y": 5(2 - ½y) + 3y = 9 .

• 5(2 - ½y) + 3y = 9
• 10 – (5/2)y + 3y = 9
• 10 – (5/2)y + (6/2)y = 9 (If you don't understand this step, learn how to add fractions . This is often, but not always, necessary for this method.)
• 10 + ½y = 9

• You know that y = -2
• One of the original equations is 4x + 2y = 8 . (You can use either equation for this step.)
• Plug in -2 instead of y: 4x + 2(-2) = 8 .

• If you end up with an equation that has no variables and isn't true (for instance, 3 = 5), the problem has no solution . (If you graphed both of the equations, you'd see they were parallel and never intersect.)
• If you end up with an equation without variables that is true (such as 3 = 3), the problem has infinite solutions . The two equations are exactly equal to each other. (If you graphed the two equations, you'd see they were the same line.)

Using the Elimination Method

• You have the system of equations 3x - y = 3 and -x + 2y = 4 .
• Let's change the first equation so that the y variable will cancel out. (You can choose x instead, and you'll get the same answer in the end.)
• The - y on the first equation needs to cancel with the + 2y in the second equation. We can make this happen by multiplying - y by 2.
• Multiply both sides of the first equation by 2, like this: 2(3x - y)=2(3) , so 6x - 2y = 6 . Now the - 2y will cancel out with the +2y in the second equation.

• Your equations are 6x - 2y = 6 and -x + 2y = 4 .
• Combine the left sides: 6x - 2y - x + 2y = ?
• Combine the right sides: 6x - 2y - x + 2y = 6 + 4 .

• You have 6x - 2y - x + 2y = 6 + 4 .
• Group the x and y variables together: 6x - x - 2y + 2y = 6 + 4 .
• Simplify: 5x = 10
• Solve for x: (5x)/5 = 10/5 , so x = 2 .

• You know that x = 2 , and one of your original equations is 3x - y = 3 .
• Plug in 2 instead of x: 3(2) - y = 3 .
• Solve for y in the equation: 6 - y = 3
• 6 - y + y = 3 + y , so 6 = 3 + y

• If your combined equation has no variables and is not true (like 2 = 7), there is no solution that will work on both equations. (If you graph both equations, you'll see they're parallel and never cross.)
• If your combined equation has no variables and is true (like 0 = 0), there are infinite solutions . The two equations are actually identical. (If you graph them, you'll see that they're the same line.)

Graphing the Equations

• The basic idea is to graph both equations, and find the point where they intersect. The x and y values at this point will give us the value of x and the value of y in the system of equations.

• Your first equation is 2x + y = 5 . Change this to y = -2x + 5 .
• Your second equation is -3x + 6y = 0 . Change this to 6y = 3x + 0 , then simplify to y = ½x + 0 .
• If both equations are identical , the entire line will be an "intersection". Write infinite solutions .

• If you don't have graph paper, use a ruler to make sure the numbers are spaced precisely apart.
• If you are using large numbers or decimals, you may need to scale your graph differently. (For example, 10, 20, 30 or 0.1, 0.2, 0.3 instead of 1, 2, 3).

• In our examples from earlier, one line ( y = -2x + 5 ) intercepts the y-axis at 5 . The other ( y = ½x + 0 ) intercepts at 0 . (These are points (0,5) and (0,0) on the graph.)
• Use different colored pens or pencils if possible for the two lines.

• In our example, the line y = -2x + 5 has a slope of -2 . At x = 1, the line moves down 2 from the point at x = 0. Draw the line segment between (0,5) and (1,3).
• The line y = ½x + 0 has a slope of ½ . At x = 1, the line moves up ½ from the point at x=0. Draw the line segment between (0,0) and (1,½).
• If the lines have the same slope , the lines will never intersect, so there is no answer to the system of equations. Write no solution .

• If the lines are moving toward each other, keep plotting points in that direction.
• If the lines are moving away from each other, move back and plot points in the other direction, starting at x = -1.
• If the lines are nowhere near each other, try jumping ahead and plotting more distant points, such as at x = 10.

Practice Problems and Answers

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Video . By using this service, some information may be shared with YouTube.

• You can check your work by plugging the answers back into the original equations. If the equations end up true (for instance, 3 = 3), your answer is correct. ⧼thumbs_response⧽ Helpful 2 Not Helpful 1
• In the elimination method, you will sometimes have to multiply one equation by a negative number in order to get a variable to cancel out. ⧼thumbs_response⧽ Helpful 1 Not Helpful 1

• These methods cannot be used if there is a variable raised to an exponent, such as x 2 . For more information on equations of this type, look up a guide to factoring quadratics with two variables. [11] X Research source ⧼thumbs_response⧽ Helpful 0 Not Helpful 0

You Might Also Like

• ↑ https://www.mathsisfun.com/definitions/system-of-equations.html
• ↑ https://calcworkshop.com/systems-equations/substitution-method/
• ↑ https://www.cuemath.com/algebra/substitution-method/
• ↑ http://tutorial.math.lamar.edu/Classes/Alg/SystemsTwoVrble.aspx
• ↑ http://www.purplemath.com/modules/systlin2.htm
• ↑ http://www.virtualnerd.com/algebra-2/linear-systems/graphing/solve-by-graphing/equations-solution-by-graphing
• ↑ https://www.khanacademy.org/math/algebra/multiplying-factoring-expression/factoring-quadratics-in-two-vari/v/factoring-quadratics-with-two-variables

About This Article

To solve systems of algebraic equations containing two variables, start by moving the variables to different sides of the equation. Then, divide both sides of the equation by one of the variables to solve for that variable. Next, take that number and plug it into the formula to solve for the other variable. Finally, take your answer and plug it into the original equation to solve for the other variable. To learn how to solve systems of algebraic equations using the elimination method, scroll down! Did this summary help you? Yes No

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Solving Equations With Two Variables

Related Pages Solving Equations More Lessons for GRE Math More Algebra Lessons

This is part of a series of lessons for the quantitative reasoning section of the GRE revised General Test. In these lessons, we will learn:

• Linear Equations in Two Variables
• Solving Simultaneous Equations or Systems of Equations
• Using the Substitution Method
• Using the Elimination Method

Linear Equation In Two Variables

A linear equation in two variables, x and y, can be written in the form ax + by = c where x and y are real numbers and a and b are not both zero.

For example, 3x + 2y = 8 is a linear equation in two variables.

A solution of such an equation is an ordered pair of numbers (x, y) that makes the equation true when the values of x and y are substituted into the equation.

For example, both (2, 1) and (0, 4) are solutions of the equation but (2, 0) is not a solution. A linear equation in two variables has infinitely many solutions.

The following video shows how to complete ordered pairs to make a solution to linear equations.

Systems of Equations or Simultaneous Equations

The following diagram shows examples of how to solve systems of equations using substitution or elimination.. Scroll down the page for more examples and solutions on how to solve systems of equations or simultaneous equations..

If another linear equation in the same variables is given, it is usually possible to find a unique solution of both equations. Two equations with the same variables are called a system of equations , and the equations in the system are called simultaneous equations . To solve a system of two equations means to find an ordered pair of numbers that satisfies both equations in the system.

There are two basic methods for solving systems of linear equations, by substitution or by elimination.

Substitution Method

In the substitution method, one equation is manipulated to express one variable in terms of the other. Then the expression is substituted in the other equation.

For example, to solve the system of equations 3x + 2y = 2 y + 8 = 3x

Isolate the variable y in the equation y + 8 = 3x to get y = 3x – 8.

Then, substitute 3x – 8 for y into the equation 3x + 2y = 2. 3x + 2 (3x – 8) = 2 3x + 6x – 16 = 2 9x – 16 = 2 9x = 18

Substitute x = 2 into y = 3x – 8.to get the value for y y = 3 (2) – 8 y = 6 – 8 = – 2

Answer: x = 2 and y = –2

How to solve simultaneous equations using substitution?

Elimination Method

In the elimination method, the object is to make the coefficients of one variable the same in both equations so that one variable can be eliminated either by adding the equations together or by subtracting one from the other.

Consider the following example: 2x + 3y = –2 4x – 3y = 14

In this example the coefficients of y are already opposites (+3 and –3). Just add the two equations to eliminate y.

To get the value of y, we need to substitute x = 2 into the equation 2x + 3y = –2 2(2) + 3y = –2 4 + 3y = –2 3y = –6 y = –2

How to solve simultaneous equations using the substitution method and elimination (or combination) method

Example of the GRE Quantitative Comparison question that involves simultaneous equations

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Unit 4: Lesson 3

Systems of equations with substitution: 2y=x+7 & x=y-4.

• Systems of equations with substitution
• Systems of equations with substitution: y=4x-17.5 & y+2x=6.5
• Systems of equations with substitution: -3x-4y=-2 & y=2x-5
• Systems of equations with substitution: 9x+3y=15 & y-x=5
• Systems of equations with substitution: y=-5x+8 & 10x+2y=-2
• Systems of equations with substitution: y=-1/4x+100 & y=-1/4x+120
• Substitution method review (systems of equations)

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• My Preferences
• My Reading List
• Basic Math & Pre-Algebra
• Study Guides
• Solving Simple Equations
• Multiplying and Dividing Using Zero
• Common Math Symbols
• Quiz: Ways to Show Multiplication and Division, Multiplying and Dividing by Zero, and Common Math Symbols
• Properties of Basic Mathematical Operations
• Quiz: Properties of Basic Mathematical Operations
• Grouping Symbols and Order of Operations
• Groups of Numbers
• Quiz: Groups of Numbers
• Ways to Show Multiplication and Division
• Order of Operations
• Quiz: Grouping Symbols and Order of Operations
• Estimating Sums, Differences, Products, and Quotients
• Quiz: Estimating Sums, Differences, Products, and Quotients
• Divisibility Rules
• Quiz: Divisibility Rules
• Factors, Primes, Composites, and Factor Trees
• Place Value
• Quiz: Factors, Primes, Composites, and Factor Trees
• Quiz: Place Value
• Using the Place Value Grid
• Quiz: Using the Place Value Grid
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• Quiz: Decimal Computation
• What Are Decimals?
• Repeating Decimals
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• Mixed Numbers
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• Quiz: Proper and Improper Fractions, Mixed Numbers, and Renaming Fractions
• What Are Fractions?
• Quiz: Factors and Multiples
• Adding and Subtracting Fractions
• Adding and Subtracting Mixed Numbers
• Quiz: Adding and Subtracting Fractions and Mixed Numbers
• Multiplying Fractions and Mixed Numbers
• Dividing Fractions and Mixed Numbers
• Quiz: Multiplying and Dividing Fractions and Mixed Numbers
• Simplifying Fractions and Complex Fractions
• Quiz: Simplifying Fractions and Complex Fractions
• Changing Fractions to Decimals
• Changing Terminating Decimals to Fractions
• Changing Infinite Repeating Decimals to Fractions
• Quiz: Changing Fractions to Decimals, Changing Terminating Decimals to Fractions, and Changing Infinite Repeating Decimals to Fractions
• Applications of Percents
• Quiz: Applications of Percents
• Changing Percents, Decimals, and Fractions
• Important Equivalents
• Quiz: Changing Percents, Decimals, and Fractions, and Important Equivalents
• Quiz: Rationals (Signed Numbers Including Fractions)
• Quiz: Integers
• Rationals (Signed Numbers Including Fractions)
• Quiz: Square Roots and Cube Roots
• Powers and Exponents
• Quiz: Powers and Exponents
• Square Roots and Cube Roots
• Quiz: Scientific Notation
• Powers of Ten
• Quiz: Powers of Ten
• Scientific Notation
• Metric System
• Converting Units of Measure
• Quiz: U.S. Customary System, Metric System, and Converting Units of Measure
• Significant Digits
• Quiz: Precision and Significant Digits
• U.S. Customary System
• Calculating Measurements of Basic Figures
• Quiz: Calculating Measurements of Basic Figures
• Quiz: Bar Graphs
• Line Graphs
• Quiz: Line Graphs
• Circle Graphs or Pie Charts
• Introduction to Graphs
• Quiz: Circle Graphs or Pie Charts
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• Quiz: Coordinate Graphs
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• Quiz: Probability
• Arithmetic Progressions
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• Quiz: Arithmetic Progressions and Geometric Progressions
• Quiz: Variables and Algebraic Expressions
• Quiz: Solving Simple Equations
• Variables and Algebraic Expressions
• Quiz: Solving Process and Key Words
• Solving Process
• Basic Math Quizzes

Solving an equation is the process of getting what you're looking for, or  solving for , on one side of the equals sign and everything else on the other side. You're really sorting information. If you're solving for  x , you must get x  on one side by itself.

Addition and subtraction equations

Some equations involve only addition and/or subtraction.

Solve for  x .

x  + 8 = 12

To solve the equation  x  + 8 = 12, you must get  x  by itself on one side. Therefore, subtract 8 from both sides.

To check your answer, simply plug your answer into the equation:

Solve for  y .

y  – 9 = 25

To solve this equation, you must get  y  by itself on one side. Therefore, add 9 to both sides.

To check, simply replace  y  with 34:

x  + 15 = 6

To solve, subtract 15 from both sides.

To check, simply replace  x  with –9 :

Notice that in each case above,  opposite operations  are used; that is, if the equation has addition, you subtract from each side.

Multiplication and division equations

Some equations involve only multiplication or division. This is typically when the variable is already on one side of the equation, but there is either more than one of the variable, such as 2  x , or a fraction of the variable, such as

In the same manner as when you add or subtract, you can multiply or divide both sides of an equation by the same number,  as long as it is not zero , and the equation will not change.

Divide each side of the equation by 3.

To check, replace  x  with 3:

To solve, multiply each side by 5.

To check, replace  y  with 35:

Or, without canceling,

Combinations of operations

Sometimes you have to use more than one step to solve the equation. In most cases, do the addition or subtraction step first. Then, after you've sorted the variables to one side and the numbers to the other, multiply or divide to get only one of the variables (that is, a variable with no number, or 1, in front of it:  x , not 2  x ).

2  x  + 4 = 10

Subtract 4 from both sides to get 2  x  by itself on one side.

Then divide both sides by 2 to get  x .

To check, substitute your answer into the original equation:

5x  – 11 = 29

Add 11 to both sides.

Divide each side by 5.

To check, replace  x  with 8:

Subtract 6 from each side.

To check, replace  x  with 9: 

Add 8 to both sides.

To check, replace  y  with –25: 

3  x  + 2 =  x  + 4

Subtract 2 from both sides (which is the same as adding –2).

Subtract  x  from both sides.

Note that 3  x  –  x  is the same as 3  x  – 1  x .

Divide both sides by 2.

To check, replace  x  with 1:

5  y  + 3 = 2  y  + 9

Subtract 3 from both sides.

Subtract 2  y  from both sides.

Divide both sides by 3.

To check, replace  y  with 2:

Sometimes you need to simplify each side (combine like terms) before actually starting the sorting process.

 Solve for  x .     

3  x  + 4 + 2 = 12 + 3

First, simplify each side.

Subtract 6 from both sides.

To check, replace  x  with 3: 

4  x  + 2  x  + 4 = 5  x  + 3 + 11

Simplify each side.

6  x  + 4 = 5  x  + 14

Subtract 4 from both sides.

Subtract 5  x  from both sides.

To check, replace  x  with 10: 

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Algebra Solutions

Our goal here is to introduce some of the equation solving techniques that may be helpful for kids in understanding Algebra.

We start here with a very simple technique.

Let us say that there is an equation x + 3 = 9 and we need to solve it. Solving an equation means that we need to know the possible values of the variables that when put into the equation will satisfy it. We present the solution next.

Explanation:.

x + 3 = 9 (original equation ) x = 9 – 3 (take the constant term to the other side so that the variable x is isolated.) x = 6

So x = 6 is the required solution to the equation.

You have an equation 3x = 9. Find x.

3x = 9 (original equation) x = 9/3 (take the constant term associated with x to the other side so that the variable x is isolated.) x = 3

So x = 3 is the required solution to the equation.

We have two simultaneous equations in two variables x and y. Find x and y. x – y = 10 --------(1) x + y = 15 --------(2)

  (25/2, 5/2).

In solving these equations, we use a simple Algebraic technique called " Substitution Method ". In this method, we evaluate one of the variable value in terms of the other variable using one of the two equations. And that value is put into the second equation to solve for the two unknown values.

The solution below will make the idea of Substitution clear.

Using 1 st equation,

x – y = 10 -----(1) x = 10 + y

Now we put this value of x into the 2 nd equation.

x + y = 15 -----(2) (10 + y) + y = 15 10 + 2y = 15 2y = 15 – 10 = 5 y = 5/2

Putting this value of y into any of the two equations will give us the value of x.

x + y = 15 x + 5/2 = 15 x = 15 – 5/2 x = 25/2

Hence (x , y) = (25/2, 5/2) is the solution to the given system of equations.

We have two simultaneous equations in two variables x and y and we need to find x and y. x – y = 10 -----(1) x + y = 15 -----(2)

 (25/2, 5/2).

In Elimination Method, our aim is to "eliminate" one variable by making the coefficients of that variable equal and then adding/subtracting the two equations, depending on the case.

In this example, we see that the coefficients of all the variable are same, i.e., 1. So if we add the two equations, the –y and the +y will cancel each other giving as an equation in only x. Let me illustrate this below.

x – y = 10 x + y = 15 2x = 25 x = 25/2

Putting the value of x into any of the two equations will give y = 5/2

Elimination Method - By Equating Coefficients:

This is another very easy and useful equation solving technique that is extensively used in Algebraic calculations. We illustrate this method through an example.

Given two equations in two variables x and y. Find the values of x and y that satisfy these equations simultaneously. 2x – y = 10 ------(1) x + 2y = 15 ------(2)

 ( 7, 4).

In this example, we see that neither the coefficients of x nor those of y are equal in the two equations. So simple addition and subtraction will not lead to a simplified equation in only one variable. However, we can multiply a whole equation with a coefficient (say we multiply equation (2) with 2) to equate the coefficients of either of the two variables.

After multiplication, we get

2x + 4y = 30 ------(2)'

Next we subtract this equation (2)’ from equation (1)

2x – y = 10 2x + 4y = 30 –5y = –20 y = 4

Putting this value of y into equation (1) will give us the correct value of x.

2x – y = 10 ------(1) 2x – 4 = 10 2x = 10 + 4 = 14 x = 14/2 = 7

Hence (x , y) =( 7, 4) gives the complete solution to these two equations.

Different Equation Types

In Algebra , sometimes you may come across equations of the form Ax + B = Cx + D where x is the variable of the equation, and A,B,C,D are coefficient values (can be both positive and negative).

In the next section, we present an example of this type of equation and learn how to solve it through simple Algebraic techniques.

How to find system of linear equations using Simple Algebric techniques

Find a value of x that satisfies this equation: 4x – 3 = 3x + 8.

We are given that

4x – 3 = 3x + 8

Separating the variables and the coefficients gives:

4x – 3x = 8 + 3

(Note: Taking a constant or a variable term to the left hand side from the right hand side (or vice versa) changes its sign as illustrated above.)

Simplifying the above equation on the L.H.S (Left Hand Side) and the R.H.S (Right Hand Side) gives

Hence x = 11 is the required solution to the above equation.

In the equation A x + B = C x + D , the coefficients A, B, C, D may also be any decimal numbers. For example, the equation could be of this form: 4x + 3.2 = 6.1x + 5.2 -- But you are not supposed to be confused with the method. There will be no change in the equation solving strategy and once you have learnt the above method, you do not need to bother about the coefficients at all.

Next we present and try to solve the examples in a more detailed step-by-step approach. Examples given next are similar to those presented above and have been shown in a way that is more understandable for kids.

Solve x + 4 = 11 to find the value of x.

x + 4 – 4 = 11 – 4 x = 7

Hence x = 7 is the solution to the given equation.

How to solve system of Linear Equations by Elimination

Solve the following system using elimination. 2x + y = 15 3x – y = 10,  (5, 5).

If we use the method of addition in solving these two equations, we can see that what we get is a simplified equation in one variable, as shown below.

2x + y = 15 ------(1) 3x – y = 10 ------(2) ______________ 5x = 25

What we are left with is a simplified equation in x alone. i.e., 5x = 25

5x/5 = 25/5 x = 5

2(5) + y = 15 10 + y = 15

Which is another equation in a single variable y.

10 + y – 10 = 15 – 10 y = 5

Hence the solution to the system of equations is (x , y) = (5, 5)

Solve the following system using Elimination x + 2y = 15 x – y = 10

 ( 35/3 , 5/( 3 )).

With a little observation, we can conclude that if we directly add these two equations, we are not going to reach any simple equation. Let us show this below.

x + 2y = 15 ------(1) x – y = 10 ------(2) ______________ 2x + y = 25

Which is another equation in 2 variables x and y. So our problem doesn’t seem to have reduced. If instead of adding the two equations directly, I multiply the entire equation (1) with – 1, and then add the resulting equation into equation (2), the +x will be cancelled out with – x as shown next.

– ( x + 2y ) = – 15 – x – 2y = – 15 ------(1’)

– x – 2y = – 15 ------(1’) x – y = + 10 ------(2) ______________ – 3y = – 5

-3y/-3=-5/-3 y = 5/3

x – 5/3 = + 10

x – 5/3 + 5/3 = 10 + 5/3 x = (30+ 5)/3 = 35/3

Hence the solution to the given system of equations is (x , y) = ( 35/3 , 5/( 3 ))

Solve the following system of linear equations using Elimination. 8x – 13y = 2 –4x + 6.5y = –2

Apparently, this system seems to be a bit complex and one might think that no cancellation of terms is possible. But a close observation and a simple multiplication can lead us in the right direction.

We are given two equations:

8x – 13y = 2 ------(1) –4x + 6.5y = –2 ------(2)

2(–4x + 6.5y ) = 2(–2) –8x + 13y = –4 ------(2’)

8x – 13y = 2 ------(1) –8x + 13y = –4 ------(2’) ______________ 0 = –2

But this is not true!! 0≠ –2

Hence the two equations constitute an inconsistent system of linear equations and thus do no have a solution (At no point do the two straight lines intersect => No solution!)

How to solve system of Linear Equations by Substitution

Solve the following system of linear equations by substitution. 2x – 2y = –2 x + y = 24,  23/2 , 25/( 2 ).

In this method of equation solving, we work out on any of the given equations for one variable value, and then substitute that value in the other equation. It gives us an equation in a single variable and we can use a single variable equation solving technique to find the value of that variable (as shown in examples above). Let us solve the given system now

2x – 2y = –2 ------(1) x + y = 24 ------(2)

Now the next question is: which equation to pick up. There is no particular criteria for this choice. One can simply choose an equation that makes the calculations simpler. E.g., in this example, the equation (2) is easier to work on.

x + y = 24 ------(2)

2(24 – y) – 2y = –2 48 – 2y – 2y = –2 48 – 4y = – 2

48 – 4y – 48 = –2 –48 –4y = –50

-4y/-4 = -50/-4 y = 50/4 = 25/2

x + 25/2 = 24

x + 25/2 - 25/2 = 24 - 25/2 x = (48 - 25)/2 = 23/2

Hence the solution to the given system of equations is (x , y) = ( 23/2 , 25/( 2 ))

Note: Next we show what happens if we substitute the value of x into the same equation that we used to compute it (equation (2) in this example)

x + y = 24 24 – y + y = 24 ∵ (x = 24 – y) 24 = 24

This is the result that we are left with. There is nothing wrong with 24 being equal to 24, but then what should we do with it? Of course we have not been looking to prove this in the first place!!

Hence we conclude that there is no point in substituting the computed value into the same equation that was used for its computation. Always use the other equation!

Solve the following system of linear equations by method of substitution. y = 24 – 4x 2x + y/2 = 12

  y = 24 – 4x.

As shown in the above example, we compute the variable value from one equation and substitute it into the other.

y = 24 – 4x ------(1) 2x + y/2 = 12 ------(2)

Here we choose equation (1) to compute the value of x. Since equation (1) is already in its most simplified form:

(Putting this value of y into equation (2) and then solving for x gives)

2x + (24-4x)/2 = 12 ------(2) (∵ y = 24 – 4x) 2x + 24/2- 4x/2 = 12 2x + 12 – 2x = 12 12 = 12

You might feel that this is the same scenario as discussed above (that of 24 = 24). But wait! You are trying to jump at a conclusion a bit too early. In the previous scenario, the result 24 = 24 had resulted because we put the variable value into the same equation that we used for its computation. Here we have not done that.

The result 12 = 12 has got something to do with the nature of the system of equations that we are given. No matter what solving technique you might be using, a solution to a system of linear equations lies at a single point where their lines intersect. In this scenario, the two lines are basically the same (one line over the other. The following figure shows this scenario.

Such a system is called a dependent system of equations. And solution to such a system is the entire line (every point on the line is a point of intersection of the two lines)

Hence the solution to the given system of equations is the entire line: y = 24 – 4x

Another possible Scenario:

Similar to this example, there exists another scenario where substitution of one variable into the 2 nd equation leads to a result similar to one shown below:

Such a scenario arises when there exists no solution to the given system of equations. I.e., when the two lines do not intersect at any point at all.

Hence in case of such a result, where your basic Math rules seem to fail, a simple conclusion is that no solution to the given system exists. Such a system of equations is called an Inconsistent system.

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