Solved Algebra Problems

Solved Algebra Problems-66
But a close observation and a simple multiplication can lead us in the right direction.

But a close observation and a simple multiplication can lead us in the right direction.

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).

(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.

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.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.Examples given next are similar to those presented above and have been shown in a way that is more understandable for kids.If we use the method of addition in solving these two equations, we can see that what we get is a simplified equation in one variable, as shown below.In this scenario, the two lines are basically the same (one line over the other. Such a system is called a dependent system of equations.And solution to such a system is the entire line (every point on the line is a point of intersection of the two lines) Hence the solution to the given system of equations is the entire line: y = 24 – 4x Another possible Scenario: Similar to this example, there exists another scenario where substitution of one variable into the 2 equation leads to a result similar to one shown below: 23 = –46 Or 5 = 34 Such a scenario arises when there exists no solution to the given system of equations.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.So if we add the two equations, the –y and the y will cancel each other giving as an equation in only x. x – y = 10 x y = 15 2x = 25 x = 25/2 Putting the value of x into any of the two equations will give y = 5/2 Hence (x , y) = (25/2, 5/2) is the solution to the given system of equations.Elimination Method - By Equating Coefficients: 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.

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