Question

Q. 33. Determine the values of and b for which
the given system of equations has infinitely many
solutions: 2x - (2a + 5y = 5 and (2b+1x-9y = 15.
Solution Given​

Answer 1

Answer:

$\large{\underline{\boxed{\sf a = - 1\: \: and \: \: b =\dfrac{5}{2}}}}$

Step-by-step explanation:

Given equation ;

• 2x - (2a + 5)y = 5
• (2b + 1)x - 9y = 15

For infinitely many solutions,

$\mathrm{ \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_1} }$

Here,

• a₁ = 2, a₂ = (2b + 1)
• b₁ = (2a + 5), b₂ = 9
• c₁ = 5, c₂ = 15

Substituting the value in above equation;

$\sf \dfrac{2}{2b + 1} = \dfrac{2a + 5}{9} = \dfrac{5}{15}$

On solving first,

$\implies \sf \frac{2}{2b + 1} = \frac{5}{15} = \frac{1}{3} \\ \\ \underline{ \bf on \: cross - multiplying :} \\ \\ \implies \sf 2b + 1 = 6 \\ \\ \implies \sf 2b = 6 - 1 \\ \\ \implies \sf 2b = 5 \\ \\ \red {\boxed{ \bf \therefore \: b = \frac{5}{2}}}$

And, on solving second part;

$\implies \sf \frac{2a + 5}{9} = \frac{5}{15} = \frac{1}{3} \\ \\ \underline{\bf on \: cross - multipling : } \\ \\ \implies \sf 3(2a + 5) = 9 \\ \\ \implies \sf 2a + 5 = \frac{9}{3} \\ \\ \implies \sf 2a + 5 = 3 \\ \\ \implies \sf 2a = 3 - 5 \\ \\ \implies \sf 2a = - 2 \\ \\ \implies \sf a = - \frac{2}{2} \\ \\ \boxed{\red{ \bf \therefore \: a = - 1}}$

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Hence, for a = (-1) and b = 5/2, the above equation has infinitely many solutions.

Answer 2

2x - (2a + 5)y = 5

(2b + 1)x - 9y = 15

___________ [GIVEN EQUATIONS]

• We have to find the value of a and b for infinitely many solutions.

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$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ = $\dfrac{c_{1}}{c_{2}}$

Here

• $a_{1}$ = 2

$a_{2}$ = 2b + 1

• $b_{1}$ = -(2a + 5)

$b_{2}$ = -9

• $c_{1}$ = 5

$c_{2}$ = 15

$\implies$ $\dfrac{2}{2b\:+\:1}$ = $\dfrac{-(2a\:+\:5)}{-9}$ = $\dfrac{5}{15}$

$\implies$ $\dfrac{2}{2b\:+\:1}$ = $\dfrac{2a\:+\:5}{9}$ = $\dfrac{5}{15}$

$\implies$ $\dfrac{2}{2b\:+\:1}$ = $\dfrac{5}{15}$

$\implies$ $\dfrac{2}{2b\:+\:1}$ = $\dfrac{1}{3}$

Cross-multiply them

$\implies$ 3 × 2 = 2b + 1

$\implies$ 6 = 2b + 1

$\implies$ 2b = 5

$\implies$ $\boxed{b \:= \:\dfrac{5}{2}}$

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Similarly..

$\implies$ $\dfrac{2a\:+\:5}{-9}$ = $\dfrac{5}{15}$

$\implies$ $\dfrac{2a\:+\:5}{9}$ = $\dfrac{1}{3}$

Cross-multiply them

$\implies$ 3(2a + 5) = 9

$\implies$ 6a + 15 = 9

$\implies$ 6a = - 6

$\implies$ $\boxed{a \:= \:- 1}$

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$\huge{\bold{a\:=\:-1}}$

and

$\huge{\bold{b\:=}}$ $\huge{\dfrac{5}{2}}$

____________ $\bold{[ANSWER]}$

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