*When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.*

The correct answer is to add Equation A and Equation B.

Just as with the substitution method, the elimination method will sometimes eliminate both variables, and you end up with either a true statement or a false statement.

And since x y = 8, you are adding the same value to each side of the first equation.

If you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables.

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.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.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.So let’s now use the multiplication property of equality first.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.$$ \begin &x 3y = -5 \color\ &\underline \end\ \begin &\underline} \text\ &-13x = 26 \end $$ 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 $(x, y) = (1, -2)$.The elimination method of solving systems of equations is also called the addition method.To solve a system of equations by elimination we transform the system such that one variable "cancels out".

## Comments Solving Problems By Elimination

## The Elimination Method

The elimination method for solving systems of linear equations uses the addition. Example. Problem. Use elimination to solve the system. x – y = −6. x + y = 8.…

## Solving by Elimination 1

Let's just do one and you'll see how it works See how these guys are the same, but with a different sign?…

## Solving systems of equations by elimination video Khan.

An old video where Sal introduces the elimination method for systems of linear equations. Let's explore a few more methods for solving systems of equations. found skills to tackle a word problem, our newly found skills in elimination.…

## 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?…

## 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.…

## Algebra 37 - Solving Systems of Equations by Elimination.

Sep 24, 2014. Systems of two equations in x and y can be solved by adding the equations to create a new equation with one variable eliminated. This new.…

## 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.…