Solving For X Practice Problems

Solving For X Practice Problems-27
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.

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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.

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.

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.

You can review your answers and change them by checking the desired letter.

Once you have finished, press "finish" and you get a table with your answers and the right answers to compare with.

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.


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