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Today I am going to talk about solving simultaneous equations. Simultaneous equations can appear in two forms: linear and linear or linear and quadratic. Luckily for you, the S.A.T only gives linear and linear, however quadratic and linear may also come but the method of solving them on the S.A.T is based on a graph, so it is a lot easier than solving it the proper way. There are two methods of solving simultaneous equations: substitution or elimination. First I shall demonstrate the substitution method.
Consider the linear equations
y = -x + 8…………………eq’n 1
4y = 3x – 24………………eq’n 2
ok, in equation 1 , y = -x + 8, and in eq’n (equation) 2 there exist a term 4y. So substitute y = -x +8 into the term 4y in eq’n 2.
y = -x + 8
4y = 3x – 24
substitute , y = -x + 8 into eq’n 2
4(-x + 8) = 3x – 24…………………now expand this
-4x + 32 = 3x – 24………………….now put like terms on one side
24 + 32 = 3x + 4x…………………...now simplify
56 = 7x……………………………...now solve for x
8 = x………………………………...now sub x into eq’n 1 to get y
y = -(8) + 8
y =0
Therefore the answer for the simultaneous equations are x = 8 and y = 0. But what this really means??? It means that if these two lines were drawn on graph paper, the answer gives the point of intersection. For these two linear equations, the point (8,0) is their point of intersection.
For more details, visit http://mathssat.blogspot.com/
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| Author: Raghav 11 May 2008 | Member Level: Gold Points : 2 |
Very Good article on simultaneous equations. I will visit your blog also.
thank u
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