*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.

Directions: Select the algebraic equation that correctly represents the given sentence. Feedback to your answer is provided in the RESULTS BOX.

Below is a math problem solver that lets you input a wide variety of math problems and it will provide the final answer for free. The version below will show you the final answer only.

Next we present and try to solve the examples in a more detailed step-by-step approach.

Examples given next are similar to those presented above and have been shown in a way that is more understandable for kids.

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.

In the next section, we present an example of this type of equation and learn how to solve it through simple Algebraic techniques. In the equation Ax B = Cx D, the coefficients A, B, C, D may also be any decimal numbers.

You'll see a button "View steps" and this takes you to the developer's site where you can purchase the full version of the solver (where you can see the steps).

Click "Show Answer" underneath the problem to see the answer.

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.

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