Solved Algebra Problems

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.

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Let us solve the given system now We are given that 2x – 2y = –2 ------(1) x y = 24 ------(2) Now the next question is: which equation to pick up. One can simply choose an equation that makes the calculations simpler.

E.g., in this example, the equation (2) is easier to work on.

Of course we have not been looking to prove this in the first place!!

Hence we conclude that there is no point in substituting the computed value into the same equation that was used for its computation. As shown in the above example, we compute the variable value from one equation and substitute it into the other.

In the previous scenario, the result 24 = 24 had resulted because we put the variable value into the same equation that we used for its computation. The result 12 = 12 has got something to do with the nature of the system of equations that we are given.

No matter what solving technique you might be using, a solution to a system of linear equations lies at a single point where their lines intersect.

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.

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.

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