# Solving Problems By Elimination

Felix may notice that now both equations have a constant of 25, but subtracting one from another is not an efficient way of solving this problem.Instead, it would create another equation where both variables are present.

Tags: Example Of A Literature Review IntroductionThesis On Thomas MooreEssay On Hysterical NeurosisA Thousand Splendid Suns Essay QuestionsHow To Solve Relationship ProblemsCover Letter Customer Service

This method for solving a pair of simultaneous linear equations reduces one equation to one that has only a single variable.

Once this has been done, the solution is the same as that for when one line was vertical or parallel.

If you multiply the second equation by −4, when you add both equations the y variables will add up to 0. Instead of multiplying one equation in order to eliminate a variable when the equations were added, you could have multiplied both equations by different numbers. Adding 4x to both sides of Equation A will not change the value of the equation, but it will not help eliminate either of the variables—you will end up with the rewritten equation 7y = 5 4x.

The correct answer is to add Equation A and Equation B. Multiplying Equation A by 5 yields 35y − 20x = 25, which does not help you eliminate any of the variables in the system.

So let’s now use the multiplication property of equality first.

## Career Essay Outline - Solving Problems By Elimination

You can multiply both sides of one of the equations by a number that will result in the coefficient of one of the variables being the opposite of the same variable in the other equation. Notice that the first equation contains the term 4y, and the second equation contains the term y.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.Use elimination when you are solving a system of equations and you can quickly eliminate one variable by adding or subtracting your equations together.$$\begin &x 3y = -5 \color\ &\underline \end\ \begin &\underline} \text\ &-13x = 26 \end$$ Now we can find: $y = -2$ Take the value for y and substitute it back into either one of the original equations.$$\begin x 3y &= -5 \ x 3\cdot(\color) &= -5\ x - 6 &= -5\ x &= 1 \end$$ The solution is $(x, y) = (1, -2)$.One child ticket costs .50 and one adult ticket costs .00. Adding or subtracting two equations in order to eliminate a common variable is called the elimination (or addition) method.Once one variable is eliminated, it becomes much easier to solve for the other one.$y 2x=6\underline=8y\: \: \: \: \; \; \; \; =16\begin \: \: \: y\: \: \: \: \: \; \; \; \; \; =2 \end$$The value of y can now be substituted into either of the original equations to find the value of x y 2x=6$$$\cdot 2x=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.Substituting the value of y = 3 in equation (i), we get 2x 3y = 11 or, 2x 3 × 3 = 11or, 2x 9 = 11 or, 2x 9 – 9 = 11 – 9or, 2x = 11 – 9or, 2x = 2 or, x = 2/2 or, x = 1Therefore, x = 1 and y = 3 is the solution of the system of the given equations. Solve 2a – 3/b = 12 and 5a – 7/b = 1 Solution: The given equations are: 2a – 3/b = 12 …………… (iv) Multiply equation (iii) by 5 and (iv) by 2, we get 10a – 15c = 60 …………… (vi) Subtracting (v) and (vi), we get or, c = 58 /-29 or, c = -2 But 1/b = c Therefore, 1/b = -2 or b = -1/2 Subtracting the value of c in equation (v), we get 10a – 15 × (-2) = 60 or, 10a 30 = 60 or, 10a 30 - 30= 60 - 30 or, 10a = 60 – 30 or, a = 30/10 or, a = 3 Therefore, a = 3 and b = 1/2 is the solution of the given system of equations. x/2 2/3 y = -1 and x – 1/3 y = 3 Solution: The given equations are: x/2 2/3 y = -1 …………… (ii) Multiply equation (i) by 6 and (ii) by 3, we get; 3x 4y = -6 …………… (iv) Solving (iii) and (iv), we get; or, y = -15/5 or, y = -3 Subtracting the value of y in (ii), we get; x - 1/3̶ × -3̶ = 3 or, x 1 = 3 or, x = 3 – 1 or, x = 2 Therefore, x = 2 and y = -3 is the solution of the equation.

## Comments Solving Problems By Elimination

• ###### Simultaneous Equatuions by Elimination, Maths First, Institute.

This method for solving a pair of simultaneous linear equations reduces one equation to one that. This method is known as the Gaussian elimination method.…

• ###### Elimination method - free math help -

Elimination method for solving systems of linear equations with examples, solutions and exercises.…

• ###### Systems of Equations by Elimination Method - ChiliMath

Use the Elimination Method to Solve Systems of Linear Equations. go directly to the six 6 worked examples to see how actual problems are being solved.…

• ###### Solve a system of equations using elimination Algebra 1. - IXL

Improve your math knowledge with free questions in "Solve a system of equations using elimination" and thousands of other math skills.…

• ###### Elimination Calculator - Solve System of Equations with.

Learn how to use elimination to solve your system of equations! Calculator shows you. Elimination Calculator. gives you. Need more problem types?…

• ###### The elimination method for solving linear systems Algebra 1.

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.…