Solving For X Problems

Solving For X Problems-27
(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.

(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.

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 = 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).

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. 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.

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) (Multiplying the entire equation (2) by 2 gives) 2(–4x 6.5y ) = 2(–2) –8x 13y = –4 ------(2’) (Adding the new equation (2’) to the equation (1) gives) 8x – 13y = 2 ------(1) –8x 13y = –4 ------(2’) ______________ 0 = –2 But this is not true!!

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.

In this example, we see that the coefficients of all the variable are same, i.e., 1.

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