Solving For X Problems

Solving For X Problems-25
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., 1.0≠ –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.

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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.

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.

i.e., 5x = 25 (Dividing this equation throughout by 5 gives) 5x/5 = 25/5 x = 5 (Putting this value of x into equation (1) gives) 2(5) y = 15 10 y = 15 Which is another equation in a single variable y.

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.

In Algebra, sometimes you may come across equations of the form Ax B = Cx D where x is the variable of the equation, and A, B, C, D are coefficient values (can be both positive and negative). S (Right Hand Side) gives x = 11 Hence x = 11 is the required solution to the above equation.We are given that y = 24 – 4x ------(1) 2x y/2 = 12 ------(2) Here we choose equation (1) to compute the value of x.Since equation (1) is already in its most simplified form: (Putting this value of y into equation (2) and then solving for x gives) 2x (24-4x)/2 = 12 ------(2) (∵ y = 24 – 4x) 2x 24/2- 4x/2 = 12 2x 12 – 2x = 12 12 = 12 You might feel that this is the same scenario as discussed above (that of 24 = 24). You are trying to jump at a conclusion a bit too early.There will be no change in the equation solving strategy and once you have learnt the above method, you do not need to bother about the coefficients at all.Next we present and try to solve the examples in a more detailed step-by-step approach.(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.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.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!!

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