Question

x upon 3 - y upon 7=7 & 4x upon 5 + y upon 2=5 by substitution method​

Answer 1

Solution :

$\bf{\orange{\underline{\bf{Given\::}}}}$

$\sf{\dfrac{x}{3} -\dfrac{y}{7} =7\:\:\:\:\:\:\:\& \:\:\:\:\:\dfrac{4x}{5} +\dfrac{y}{2} =5}$

$\bf{\orange{\underline{\bf{To\:find\::}}}}$

The value of x and y.

$\bf{\orange{\underline{\bf{Explanation\::}}}}$

$\bullet\sf{\dfrac{x}{3} -\dfrac{y}{7} =7..........................(1)}\\\\\bullet\sf{\dfrac{4x}{5} +\dfrac{y}{2} =5..........................(2)}$

$\underline{\underline{\bf{Using\:Substitution\:Method\::}}}}}$

From equation (1),we get;

$\longrightarrow\sf{\dfrac{x}{3} -\dfrac{y}{7} =7}\\\\\\\longrightarrow\sf{\dfrac{7x-3y}{21} =7}\\\\\\\longrightarrow\sf{7x-3y=147}\\\\\\\longrightarrow\sf{7x=147+3y}\\\\\\\longrightarrow\sf{x=\dfrac{147+3y}{7} ......................(3)}$

Putting the value of x in equation (2),we get;

$\longrightarrow\sf{\dfrac{4x}{5} +\dfrac{y}{2} =5}\\\\\\\longrightarrow\sf{\dfrac{8x+5y}{10} =5}\\\\\\\longrightarrow\sf{8x+5y=50}\\\\\\\longrightarrow\sf{8\bigg(\dfrac{147+3y}{7}\bigg) +5y=50}\\\\\\\longrightarrow\sf{\dfrac{1176+24y}{7} +5y=50}\\\\\\\longrightarrow\sf{1176+24y+35y=350}\\\\\\\longrightarrow\sf{1176+59y=350}\\\\\\\longrightarrow\sf{59y=350-1176}\\\\\\\longrightarrow\sf{59y=-826}\\\\\\\longrightarrow\sf{y=-\cancel{\dfrac{826}{59} }}$

$\longrightarrow\sf{\pink{y=-14}}$

Putting the value of y in equation (3),we get;

$\longrightarrow\sf{x=\dfrac{147+3(-14)}{7} }\\\\\\\longrightarrow\sf{x=\dfrac{147+(-42)}{7} }\\\\\\\longrightarrow\sf{x=\dfrac{147-42}{7} }\\\\\\\longrightarrow\sf{x=\cancel{\dfrac{105}{7} }}\\\\\\\longrightarrow\sf{\pink{x=15}}$

Thus;

The value of x is 15 and y is -14 .

Answer 2

Answer:-

Given :

$\frac{x}{3} - \frac{y}{7} = 7$

Taking LCM,

$\frac{7x - 3y}{21} = 7 \\ \\ → 7x - 3y = 7(21) \\ \\ → 7x = 147 + 3y \\ \\ → x = \frac{147 + 3y}{7} \: \: - \: equation \: (1).$

And,

$\frac{4x}{5} + \frac{y}{2} = 5 \\ \\ → \frac{8x + 5y}{10} = 5 \\ \\→ 8x + 5y = 50$

Substitute the value of "x" here.

→ $8( \frac{147 + 3y}{7} ) + 5y = 50 \\ \\ → \frac{1176 + 24y + 35y}{7} = 50 \\ \\→ 1176 + 59y = 350 \\ \\ → 59y = 350 - 1176 \\ \\ → 59y = - 826 \\ \\ → y = \frac{ - 826}{59} \\ \\→ y = - 14$

Substitute "y" value in equation (1).

$→ x = \frac{147 + 3y}{7} \\ \\→ x = \frac{147 + 3( - 14)}{7} \\ \\→ x = \frac{147 - 42}{7} \\ \\→ x = \frac{105}{7} \\ \\ → x = 15$

Therefore , the values of x and y are 15 & - 14.