Solving For X Practice Problems

Solving For X Practice Problems-79
(Subtracting 10 from both sides of the equation gives) 10 y – 10 = 15 – 10 y = 5 Hence the solution to the system of equations is (x , y) = (5, 5) With a little observation, we can conclude that if we directly add these two equations, we are not going to reach any simple equation. x 2y = 15 ------(1) x – y = 10 ------(2) ______________ 2x y = 25 Which is another equation in 2 variables x and y. 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.(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.If you're seeing this message, it means we're having trouble loading external resources on our website.

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5z 5 = 3z 6 11 5z 5 = 3z 17 5z = 3Z 17 – 5 5z – 3z = 12 2z = 12 z = 6 5.

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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.

Elimination Method - By Equating Coefficients: 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.

And that value is put into the second equation to solve for the two unknown values.

Solving For X Practice Problems

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.

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This is another very easy and useful equation solving technique that is extensively used in Algebraic calculations. 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.

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