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.

The solution below will make the idea of Substitution clear. 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. 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.

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.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. 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 Hence (x , y) = (25/2, 5/2) is the solution to the given system of equations.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 (Since y and –y cancel out each other) What we are left with is a simplified equation in x alone.Then the following equation can represent this problem: 17 x = 68 We can subtract 17 from both sides of the equation to find the value of x. Let's look at some examples of writing algebraic equations.68 - 17 = x Answer: x = 51, so Jeanne needs to buy the game. Example 1: Write each sentence as an algebraic equation.(Multiplying equation (1) with – 1 on both sides of the equality gives) – ( x 2y ) = – 15 – x – 2y = – 15 ------(1’) (Adding the new equation (1’) to the equation (2) gives) – x – 2y = – 15 ------(1’) x – y = 10 ------(2) ______________ – 3y = – 5 (Dividing on both sides of the equation by – 3) -3y/-3=-5/-3 y = 5/3 (Putting this value of y into equation (2) gives) x – 5/3 = 10 (Adding 5/3 to both sides of the equation gives) 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 )) 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.Or click the "Show Answers" button at the bottom of the page to see all the answers at once.If you need assistance with a particular problem, click the "step-by-step" link for an in depth solution. At Wyzant, connect with algebra tutors and math tutors nearby. Find online algebra tutors or online math tutors in a couple of clicks. Living Environment, US and Global History, Algebra Core...

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