**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.