Question

5p+q=2 6p-3q=1 slove this by substitution method, elimination method, cross-multiplication method

Answer 1

$\huge\underline\mathrm{GIVEN:-}$

• 5p+q=2
• 6p-3q=1

$\huge\underline\mathrm{TO\:FIND:-}$

Find the value of p and q by Substitution, Elimination & Cross-Multiplication method ?

$\huge\underline\mathrm{SOLUTION:-}$

$\normalsize{\boxed{\underline{\underline{\bf{\red{SUBSTITUTION\: METHOD:-}}}}}}$

●We have :-

5p+q=2 ....1)

6p-3q=1....2)

Taking equation 1)

$\sf {\mapsto q = 2-5p}$

On putting the value of q in equation 2), we get

$\sf {\mapsto 6p - 3(2-5p) = 1}$

$\sf {\mapsto 6p - 6 + 15p = 1}$

$\sf {\mapsto 21p = 1+6}$

$\sf {\mapsto 21p = 7}$

$\sf {\mapsto p = \cancel\dfrac{7}{21}}$

$\large\sf\red {\longmapsto \: \star \: p = \frac{1}{3}\: \star}$

Put the value of p in equation 1)

$\sf {\mapsto 5\times \frac{1}{3} + q = 2}$

$\sf {\mapsto \frac{5}{3} + q = 2}$

$\sf {\mapsto \frac{5 + 3q}{3} = 2}$

$\sf {\mapsto 5 + 3q = 3\times 2}$

$\sf{ \mapsto 3q = 6-5}$

$\large\sf \red{\mapsto \: \star \:q = \frac{1}{3}\: \star}$

______________

$\normalsize{\boxed{\underline{\underline{\bf{\red{ELIMINATION\: METHOD:-}}}}}}$

To Eliminate both the equations, the coefficients of both the equations should be equal

$\sf{5p + q = 2}$(Multiplying by 3)

$\sf{15p + 3q = 6}$

Taking equation 1) & 2), we get

$\large\sf{15p + \cancel{3q} = 6}$

$\large\underline\mathsf{6p - \cancel{3q}= 1}$

$\sf{21p \:\:= \:\:7}$

$\sf {\mapsto p = \cancel\dfrac{7}{21}}$

$\large\sf\red {\mapsto \: \star \:p = \frac{1}{3}\star}$

Put the value of p in equation 1)

$\sf {\mapsto 5\times \frac{1}{3} + q = 2}$

$\sf {\mapsto \frac{5}{3} + q = 2}$

$\sf {\mapsto \frac{5 + 3q}{3} = 2}$

$\sf {\mapsto 5 + 3q = 3\times 2}$

$\sf{\mapsto 3q = 6-5\: }$

$\large\sf \red{\mapsto \: \star \:q = \frac{1}{3}\: \star}$

_______________

$\normalsize{\boxed{\underline{\underline{\bf{\red{CROSS-MULTIPLICATION\:METHOD:-}}}}}}$

$\sf{p_1 = 5 }$⠀⠀$\sf{ q_1 = 1}$ ⠀$\sf{1 = -2}$

$\sf{p_2 = 6}$ ⠀$\sf{q_2 = -3}$⠀$\sf{1 = -1}$

p⠀⠀⠀⠀q⠀⠀⠀⠀1

5⠀⠀⠀⠀1⠀⠀⠀⠀–2

6⠀⠀⠀⠀-3⠀⠀⠀ –1

✪  For (p) cross -multiplying between q and 1 ✪ 

✪  For (q) cross-multiplying between p and 1 ✪ 

✪  For (1) cross-multiplying between p and q ✪ 

$\large\sf{\frac{p}{-1-6}}$⠀$\large\sf{\frac{q}{-5-(-12)}}$⠀$\large\sf{\frac{1}{-15-6}}$⠀

$\large\sf{\frac{p}{-7}}$⠀⠀⠀$\large\sf{\frac{-q}{7}}$⠀⠀⠀$\large\sf{\frac{1}{-21}}$

Taking p and 1

$\large\sf{\frac{p}{-7} =\frac{1}{-21}}$

$\sf{p = \cancel\dfrac{-7}{-21}}$

$\sf\red{\longmapsto\:\star \:p = \frac{1}{3}\star}$

Thus, the value of p is $\bf{\frac{1}{3}}$

Taking q and 1

$\sf{\frac{q}{-7} =\frac{1}{-21}}$

$\sf{q = \cancel\dfrac{-7}{-21}}$

$\sf\red{ \longmapsto\:\star\:q = \frac{1}{3}\:\star}$

Thus, the value of q is $\bf{\frac{1}{3}}$

______________

$\large\underline\mathrm{FINAL\: ANSWER:-}$

$\Large{\boxed{\boxed{\bf{\red{P = \frac{1}{3}}}}}}$

$\Large{\boxed{\boxed{\bf{\red{q = \frac{1}{3}}}}}}$