# How do you solve the simultaneous equations 2x - 3y=11 and  5x+2y=18?

Mar 25, 2018

$x = 4 , y = - 1$

#### Explanation:

There Are Actually 4 methods of solving this.

We have,

$2 x - 3 y = 11$.................................(i)

and, $5 x + 2 y = 18$.........................(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,

$4 x - 6 y = 22$...........................................(iii)

Now, Multiply the eq(ii) with 3.

So we get,

$15 x + 6 y = 54$.......................................(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.]

We get,

$\textcolor{w h i t e}{\times x} 4 x \cancel{- 6 y} + 15 x \cancel{+ 6 y} = 22 + 54$

$\Rightarrow 19 x = 76$

$\Rightarrow x = 4$

So, We Got $x$.

Now, Substitute $x = 4$ in eq(i).

So, We get,

$\textcolor{w h i t e}{\times x} 2 \cdot 4 - 3 y = 11$

$\Rightarrow 8 - 3 y = 11$

$\Rightarrow - 3 y = 11 - 8$

$\Rightarrow - 3 y = 3$

$\Rightarrow y = - 1$

Thus, The Answer is $x = 4 , y = - 1$.

ii) Substitution Method

First Express one variable with the help of another.

From eq(i),

$\textcolor{w h i t e}{\times x} 2 x - 3 y = 11$

$\Rightarrow 2 x = 11 + 3 y$ [Transpose $3 y$ to R.H.S]

$\Rightarrow x = \frac{11 + 3 y}{2}$.....................................(v)

Now substitute this value in eq (ii).

So, We get,

$\textcolor{w h i t e}{\times x} 5 \left(\frac{11 + 3 y}{2}\right) + 2 y = 18$

$\Rightarrow \frac{55 + 15 y}{2} + 2 y = 18$

$\Rightarrow \frac{55 + 15 y + 4 y}{2} = 18$ [Simplify]

$\Rightarrow 55 + 19 y = 36$

$\Rightarrow 19 y = 36 - 55$

$\Rightarrow y = - \frac{19}{19} = - 1$

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,

$\textcolor{w h i t e}{\times x} 2 x - 3 \cdot \left(- 1\right) = 11$

$\Rightarrow 2 x + 3 = 11$

$\Rightarrow 2 x = 8$

$\Rightarrow x = 4$

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),

$\textcolor{w h i t e}{\times x} 2 x - 3 y = 11$

$\Rightarrow 2 x = 11 + 3 y$ [Transpose $3 y$ to R.H.S]

$\Rightarrow x = \frac{11 + 3 y}{2}$.....................................(vi)

From eq(ii),

$\textcolor{w h i t e}{\times x} 5 x + 2 y = 18$

$\Rightarrow 5 x = 18 - 2 y$ [Transpose $3 y$ to R.H.S]

$\Rightarrow x = \frac{18 - 2 y}{5}$.....................................(vii)

Now, From eq(vi) and eq(vii), We get,

$\textcolor{w h i t e}{\times x} \frac{11 + 3 y}{2} = \frac{18 - 2 y}{5}$

$\Rightarrow 5 \left(11 + 3 y\right) = 2 \left(18 - 2 y\right)$ [Cross Multiply]

$\Rightarrow 55 + 15 y = 36 - 4 y$

$\Rightarrow 19 y = 36 - 55$

$\Rightarrow 19 y = - 19$

$\Rightarrow y = - 1$

Now, Substitute this in eq(i) or eq(ii).

I'm going with eq(i) here too.

So, We get,

$\textcolor{w h i t e}{\times x} 2 x - 3 \cdot \left(- 1\right) = 11$

$\Rightarrow 2 x + 3 = 11$

$\Rightarrow 2 x = 8$

$\Rightarrow x = 4$

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 ($a x + b y + c = 0$).

So, From eq(i),

$2 x - 3 y - 11 = 0$.......................................(viii)

and $5 x + 2 y - 18 = 0$................................(ix)

Now, Use The Formula.

$\textcolor{w h i t e}{\times x} \frac{x}{{b}_{1} {c}_{2} - {b}_{2} {c}_{1}} = \frac{y}{-} \left({a}_{1} {c}_{2} - {a}_{2} {c}_{1}\right) = \frac{1}{{a}_{1} {b}_{2} - {a}_{2} {b}_{1}}$

$\Rightarrow \frac{x}{\left(- 3 \times - 18\right) - \left(2 \times - 11\right)} = \frac{y}{-} \left(\left(2 \times - 18\right) - \left(5 \times - 11\right)\right) = \frac{1}{\left(2 \times 2\right) - \left(5 \times - 3\right)}$

$\Rightarrow \frac{x}{54 + 22} = \frac{y}{-} \left(- 36 + 55\right) = \frac{1}{4 + 15}$

$\Rightarrow \frac{x}{76} = \frac{y}{-} 19 = \frac{1}{19}$

So, $\frac{x}{76} = \frac{1}{19} \Rightarrow x = \frac{76}{19} = 4$

And, $\frac{y}{-} 19 = \frac{1}{19} \Rightarrow y = \frac{19}{-} 19 = - 1$

So, The Answer is $x = 4 , y = - 1$.

Hence Explained.