How To Solve An Algebra Problem

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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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.

(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.

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.

x y = 24 ------(2) (we find the value of x in terms of y) x = 24 –y (Next we put this value of x into equation (1)) 2(24 – y) – 2y = –2 48 – 2y – 2y = –2 48 – 4y = – 2 (Subtracting 48 from both sides of the equation gives) 48 – 4y – 48 = –2 –48 –4y = –50 (Dividing on both sides of the equation by – 4) -4y/-4 = -50/-4 y = 50/4 = 25/2 (Putting this value of y into equation (2) and then solving for x gives) x 25/2 = 24 (Subtracting 25/2 from both sides of the equation gives) x 25/2 - 25/2 = 24 - 25/2 x = (48 - 25)/2 = 23/2 Hence the solution to the given system of equations is (x , y) = ( 23/2 , 25/( 2 )) Note: Next we show what happens if we substitute the value of x into the same equation that we used to compute it (equation (2) in this example) x y = 24 24 – y y = 24 ∵ (x = 24 – y) 24 = 24 This is the result that we are left with.

There is nothing wrong with 24 being equal to 24, but then what should we do with it?

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.

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