Solved Algebra Problems

Solved Algebra Problems-86
However, we can multiply a whole equation with a coefficient (say we multiply equation (2) with 2) to equate the coefficients of either of the two variables.After multiplication, we get 2x 4y = 30 ------(2)' Next we subtract this equation (2)’ from equation (1) 2x – y = 10 2x 4y = 30 –5y = –20 y = 4 Putting this value of y into equation (1) will give us the correct value of x.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.

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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.

We are given that y = 24 – 4x ------(1) 2x y/2 = 12 ------(2) Here we choose equation (1) to compute the value of x.

Since equation (1) is already in its most simplified form: (Putting this value of y into equation (2) and then solving for x gives) 2x (24-4x)/2 = 12 ------(2) (∵ y = 24 – 4x) 2x 24/2- 4x/2 = 12 2x 12 – 2x = 12 12 = 12 You might feel that this is the same scenario as discussed above (that of 24 = 24). You are trying to jump at a conclusion a bit too early.

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.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.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.But a close observation and a simple multiplication can lead us in the right direction.We are given two equations: 8x – 13y = 2 ------(1) –4x 6.5y = –2 ------(2) (Multiplying the entire equation (2) by 2 gives) 2(–4x 6.5y ) = 2(–2) –8x 13y = –4 ------(2’) (Adding the new equation (2’) to the equation (1) gives) 8x – 13y = 2 ------(1) –8x 13y = –4 ------(2’) ______________ 0 = –2 But this is not true!!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.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).

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