WARNING! VERY LONG ANSWER!
There Are Actually 4 methods of solving this.
We have,
#2x - 3y = 11#.................................(i)
and, #5x + 2y = 18#.........................(ii)
i) Elimination Method
First choose which variable you want to eliminate.
I'm going with #y#.
So, Multiply the eq(i) with 2 first.
It will turn into,
#4x - 6y = 22#...........................................(iii)
Now, Multiply the eq(ii) with 3.
So we get,
#15x + 6y = 54#.......................................(iv)
[The main reason for doing this is to have the coefficient of #y# as
same the LCM of the two #y# coefficients of the two given equations.]
Now, Add eq(iii) and eq(iv).
We get,
#color(white)(xxx)4x cancel(- 6y) + 15x cancel(+ 6y) = 22 + 54#
#rArr 19x = 76#
#rArr x = 4#
So, We Got #x#.
Now, Substitute #x = 4# in eq(i).
So, We get,
#color(white)(xxx) 2*4 - 3y =11#
#rArr 8 - 3y = 11#
#rArr -3y = 11 - 8#
#rArr -3y = 3#
#rArr y = -1#
Thus, The Answer is #x = 4, y = -1#.
ii) Substitution Method
First Express one variable with the help of another.
From eq(i),
#color(white)(xxx)2x - 3y = 11#
#rArr 2x = 11 + 3y# [Transpose #3y# to R.H.S]
#rArr x = (11 + 3y)/2#.....................................(v)
Now substitute this value in eq (ii).
So, We get,
#color(white)(xxx)5((11 + 3y)/2) + 2y = 18#
#rArr (55 + 15y)/2 + 2y = 18#
#rArr (55 + 15y + 4y)/2 = 18# [Simplify]
#rArr 55 + 19y = 36#
#rArr 19y = 36 - 55#
#rArr y = -19/19 = -1#
So, We got the value of #y#.
Now, Substitute this in eq(i) or eq(ii).
I'm going with eq(i).
So, We get,
#color(white)(xxx)2x - 3 * (-1) = 11#
#rArr 2x +3 = 11#
#rArr 2x = 8#
#rArr x = 4#
So, The Answer is #x = 4, y = -1#.
iii) Comparison Method
First we express one of the variable with another in both the equations. As the variable is same, the value will be same for this particular system; so we can get a linear equation, which we can solve to find the other variable.
So,
From eq(i),
#color(white)(xxx)2x - 3y = 11#
#rArr 2x = 11 + 3y# [Transpose #3y# to R.H.S]
#rArr x = (11 + 3y)/2#.....................................(vi)
From eq(ii),
#color(white)(xxx)5x + 2y = 18#
#rArr 5x = 18 - 2y# [Transpose #3y# to R.H.S]
#rArr x = (18 - 2y)/5#.....................................(vii)
Now, From eq(vi) and eq(vii), We get,
#color(white)(xxx)(11 + 3y)/2 = (18 - 2y)/5#
#rArr 5(11 + 3y) = 2(18 - 2y)# [Cross Multiply]
#rArr 55 + 15y = 36 - 4y#
#rArr 19y = 36 - 55#
#rArr 19y = -19#
#rArr y = -1#
Now, Substitute this in eq(i) or eq(ii).
I'm going with eq(i) here too.
So, We get,
#color(white)(xxx)2x - 3 * (-1) = 11#
#rArr 2x +3 = 11#
#rArr 2x = 8#
#rArr x = 4#
So, The Answer is still same, #x = 4, y = -1#.
iv) Cross - Multiplication Method
The Most Tough But Most Helpful method.
First Convert The Equations in General Form (#ax + by + c = 0#).
So, From eq(i),
#2x - 3y - 11 = 0#.......................................(viii)
and #5x + 2y - 18 = 0#................................(ix)
Now, Use The Formula.
#color(white)(xxx)x/(b_1c_2 - b_2c_1) = y/-(a_1c_2 - a_2c_1) = 1/(a_1b_2 - a_2b_1)#
#rArr x/((-3xx-18) - (2 xx -11)) = y/-((2xx-18) - (5 xx -11)) = 1/((2 xx 2) - (5 xx -3))#
#rArr x/(54 + 22) = y/-(-36 + 55) = 1/(4 + 15)#
#rArr x/76 = y/-19 =1/19#
So, #x/76 = 1/19 rArr x = 76/19 = 4#
And, #y/-19 = 1/19 rArr y = 19/-19 = -1#
So, The Answer is #x = 4, y = -1#.
Hence Explained.
