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Solve the following pair of linear equations by the substitution and cross-multiplication methods: $$8x + 5y = 9$$ $$3x + 2y = 4$$

Ans.(x + y = 14\)...........(i)
=> $$x – y = 4$$..............(ii)

Using equation (ii)
=> $$x = 4 + y$$......................(iii)

On substituting (iii) in (i):
=>$$4 + 2y = 14$$
=>$$2y = 10$$
=>$$y = 5$$...........(iv)

On substituting (iv) in (i):
=>$$x + 5 = 14$$
=>$$x = 14 - 5$$
=>$$x = 9$$

So, the value of $$y = 5$$ and $$x = 9$$.
$$x - 3y – 7 = 0,3x - 3y – 15 = 0$$

On comparing quations with general form of equation $$ax^2 + bx + c$$
$$a_1 = 1,$$
$$b_1 = -3,$$
$$c_1 = -7,$$
$$a_2 = 3,$$
$$b_2 = -3,$$
$$c_2 = -15$$

Here $${{a_1}\over {a_2}} \ne {{b_1}\over {b_2}}$$
=>$${{1}\over {3}} \ne {{-3}\over {-3}}$$

So, these lines have a unique solution.

We can find the unique solution using cross multiplication:

=>$${{x}\over {(-3)(-15)-(-3)(-7)}} = {{y}\over {(-7)(3)-(-15)(1)}} = {{1}\over {(-3)(1)-(-3)(3)}}$$
=>$${{x}\over {45 - 21}} = {{y}\over {-21 + 15}} = {{1}\over {-3 + 9}}$$
=>$${{x}\over {24}} = {{y}\over {-6}} = {{1}\over {6}}$$

=> x = 4 and y = -1.