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# Solve the following pair of linear equations by the elimination method and the substitution method: (i) $$x + y = 5,2x – 3y = 4$$ (ii) $$3x + 4y = 10, 2x – 2y = 2$$ (iii) $$3x - 5y – 4 = 0, 9x = 2y + 7$$ (iv) $${{x}\over{2}} + {{2y}\over{3}} = -1, x - {{y}\over{3}} = 3$$

Ans.(i) $$x + y = 5$$ … (1)
$$2x – 3y = 4$$ … (2)

Elimination method:

On multiplying equation (1) by 2, we get equation (3)

$$2x + 2y = 10$$ … (3)
$$2x - 3y = 4$$ … (2)

On subtracting equation (2) from (3), we get
$$5y = 6$$
=> $$y = {{6}\over{5}}$$

On putting value of y in (1), we get
$$x + {{6}\over{5}} = 5$$
$$x = 5 - {{6}\over{5}} = {{19}\over{5}}$$

Therefore, x = $${{6}\over{5}}$$ and y = $${{19}\over{5}}$$

Substitution method:
$$x + y = 5$$ … (1)
$$2x - 3y = 4$$ … (2)

From equation (1), we get,
$$x = 5 - y$$

On putting this in equation (2), we get
$$2 (5 - y) - 3y = 4$$
=>$$10 - 2y - 3y = 4$$
=>$$5y = 6$$
=>$$y = {{6}\over{5}}$$

On putting value of y in (1), we get
$$x = 5 - {{6}\over{5}} = {{19}\over{5}}$$

Therefore, y = $${{6}\over{5}}$$ and x = $${{19}\over{5}}$$

(ii) $$3x + 4y = 10$$ … (1)
$$2x – 2y = 2$$ … (2)

Elimination method:

On multiplying equation (2) by 2, we get equation (3)

$$4x - 4y = 4$$ … (3)
$$3x + 4y = 10$$ … (1)

On adding equation (1) and (3), we get
$$7x = 14 => x = 2$$

On putting value of x in (1), we get
$$3 (2) + 4y = 10$$
$$4y = 10 – 6 = 4$$
$$=> y = 1$$

Therefore, x = $$2$$ and y = $$1$$

Substitution method:
$$3x + 4y = 10$$ … (1)
$$2x - 2y = 2$$ … (2)

From equation (2), we get,
$$2x = 2 + 2y$$
=> $$x = 1 + y$$..........(3)

On putting this in equation (1), we get
$$3 (1 + y) + 4y = 10$$
=>$$3 + 3y + 4y = 10$$
=>$$7y = 7 => y = 1$$

On putting value of y in (3), we get
$$x = 1 + 1 = 2$$

Therefore, x = $$2$$ and y = $$1$$

(iii) $$3x - 5y – 4 = 0$$ … (1)
$$9x = 2y + 7$$… (2)

Elimination method:

Multiplying (1) by 3, we get (3)

$$9x - 15y – 12 = 0$$… (3)
$$9x - 2y – 7 = 0$$… (2)

On subtracting (2) from (3), we get
$$-13y – 5 = 0$$
=>$$-13y = 5$$
=> $$y = {{-5} \over {13}}$$

On putting value of y in (1), we get
$$3x – 5({{-5} \over {13}}) - 4 = 0$$
=>$$3x = 4 - {{25} \over {13}}= {{52 - 25} \over {13}} = {{27} \over {13}}$$
=> $$x = {{27} \over {13(3)}} = {{9} \over {13}}$$
Therefore, x = $${{9} \over {13}}$$ and y = $${{-5} \over {13}}$$

Substitution Method:
$$3x - 5y – 4 = 0$$ … (1)
$$9x = 2y + 7$$… (2)

From equation (1), we can say that
$$3x = 4 + 5y => x = {{4 + 5y} \over {3}}$$

On putting this in equation (2), we get
$$9({{4 + 5y} \over {3}}) - 2y = 7$$
=>$$12 + 15y - 2y = 7$$
=>$$13y = -5$$
$$y = {{-5} \over {13}}$$

On putting value of y in (1), we get
$$3x – 5 ({{-5} \over {13}})= 4$$
=>$$3x = 4 - {{25} \over {13}} = {{52 - 25} \over {13}} = {{27} \over {13}}$$
=>$$x = {{27} \over {13(3)}} = {{9} \over {13}}$$

Therefore, x = $${{9} \over {13}}$$ and y = $${{-5} \over {13}}$$

(iv)$${{x}\over{2}} + {{2y}\over{3}} = -1$$ … (1)
$$x - {{y}\over{3}} = 3$$… (2)

Elimination method:

On multiplying equation (2) by 2, we get (3)

$$2x - {{2y}\over{3}} = 6$$… (3)
$${{x}\over{2}} + {{2y}\over{3}} = -1$$… (1)

Adding (3) and (1), we get

$${{5x}\over{2}} = 5 => x = 2$$

On putting value of x in (2), we get
$$2 - {{y}\over{3}} = 3$$
=>$$y = -3$$

Therefore, x = 2 and y = -3

Substitution method:

$${{x}\over{2}} + {{2y}\over{3}} = -1$$ … (1)
$$x - {{y}\over{3}} = 3$$… (2)

From equation (2), we can say that
=> $$x = 3 + {{y}\over{3}} = {{9 + y}\over{3}}$$

Putting this in equation (1), we get
=>$${{9 + y}\over{6}} + {{2y}\over{3}} = -1$$
=>$${{9 + y + 4y}\over{6}} = -1$$
=>$$5y + 9 = -6$$
=>$$5y = -15$$
=>$$y = -3$$

Putting value of y in (1), we get
=>$${{x}\over{2}} + {{2(-3)}\over{3}} = -1$$
=>$$x = 2$$

So, the value of $$y = -3$$ and $$x = 2$$.