Question

x+20=3y/2+10=2z+5=110-(y.+z)​

Answer 1

$\sf{x + 20 = \frac{3y}{2} + 10}$

$\sf{Multiplying \ 2 \ on \ both \ side}$

$\sf{2x + 40 = 3y + 20}$

$\sf{2x - 3y = 20 - 40}$

$\sf{2x - 3y = - 20 .... (1)}$

$\sf{\frac{3y}{2} + 10 = 2z + 5}$

$\sf{Multiplying \ 2 \ on \ both \ side}$

$\sf{3y + 20 = 4z + 10}$

$\sf{3y - 4z = 10 - 20}$

$\sf{3y - 4z = - 10 .... (2) }$

$\sf{2z + 5 = 110 - (y + z)}$

$\sf{2z + 4 = 110 - y - z}$

$\sf{y + 3z = 105 .... (3)}$

$\sf{Multiplying \ (2) \ by \ 1 \ eq (3) \ by \ 3}$

$\sf{also \ Subtracting \ eq(2) \ from \ eq(3)}$

$\sf{3y + 9z = 315}$

$\underline{\sf{3y - 4z = - 10}}$

$\sf{13z = 325}$

${\sf{z = \frac{325}{13} = 25 }}$

$\fbox{\sf{z = 25}}$

$\sf{Substituting \ the \ value \ of \ z = 25 \ in \ eq(2) }$

$\sf{3y - 4z = - 10}$

$\sf{3y - 4(25) = - 10}$

$\sf{3y - 100= - 10}$

$\sf{3y = - 10 + 100}$

$\sf{3y = 90}$

$\sf{y = \frac{90}{3}}$

$\fbox{\sf{y = 30}}$

$\sf{Substituting \ the \ value \ of \ y = 30 \ in \ eq(1) }$

$\sf{2x - 3y = - 20}$

$\sf{2x - 3(30) = - 20}$

$\sf{2x - 90 = - 20}$

$\sf{2x = - 20 + 90}$

$\sf{2x = 70}$

$\sf{x = \frac{70}{2}}$

$\fbox{\sf{x = 35}}$

$\sf\green{Hence \ x = 35}$

$\sf\green{y = 30}$

$\sf\green{z = 25}$

#NAWABZAADI

Answer 2

Answer:

\sf{x + 20 = \frac{3y}{2} + 10}x+20=

2

3y

+10

\sf{Multiplying \ 2 \ on \ both \ side}Multiplying 2 on both side

\sf{2x + 40 = 3y + 20}2x+40=3y+20

\sf{2x - 3y = 20 - 40}2x−3y=20−40

\sf{2x - 3y = - 20 .... (1)}2x−3y=−20....(1)

\sf{\frac{3y}{2} + 10 = 2z + 5}

2

3y

+10=2z+5

\sf{Multiplying \ 2 \ on \ both \ side}Multiplying 2 on both side

\sf{3y + 20 = 4z + 10}3y+20=4z+10

\sf{3y - 4z = 10 - 20}3y−4z=10−20

\sf{3y - 4z = - 10 .... (2) }3y−4z=−10....(2)

\sf{2z + 5 = 110 - (y + z)}2z+5=110−(y+z)

\sf{2z + 4 = 110 - y - z}2z+4=110−y−z

\sf{y + 3z = 105 .... (3)}y+3z=105....(3)

\sf{Multiplying \ (2) \ by \ 1 \ eq (3) \ by \ 3}Multiplying (2) by 1 eq(3) by 3

\sf{also \ Subtracting \ eq(2) \ from \ eq(3)}also Subtracting eq(2) from eq(3)

\sf{3y + 9z = 315}3y+9z=315

\underline{\sf{3y - 4z = - 10}}

3y−4z=−10

\sf{13z = 325}13z=325

\begin{gathered}{\sf{z = \frac{325}{13} = 25 }} \\ \end{gathered}

z=

13

325

=25

\fbox{\sf{z = 25}}

z = 25

\sf{Substituting \ the \ value \ of \ z = 25 \ in \ eq(2) }Substituting the value of z=25 in eq(2)

\sf{3y - 4z = - 10}3y−4z=−10

\sf{3y - 4(25) = - 10}3y−4(25)=−10

\sf{3y - 100= - 10}3y−100=−10

\sf{3y = - 10 + 100}3y=−10+100

\sf{3y = 90}3y=90

\begin{gathered}\sf{y = \frac{90}{3}} \\ \end{gathered}

y=

3

90

\fbox{\sf{y = 30}}

y = 30

\sf{Substituting \ the \ value \ of \ y = 30 \ in \ eq(1) }Substituting the value of y=30 in eq(1)

\sf{2x - 3y = - 20}2x−3y=−20

\sf{2x - 3(30) = - 20}2x−3(30)=−20

\sf{2x - 90 = - 20}2x−90=−20

\sf{2x = - 20 + 90}2x=−20+90

\sf{2x = 70}2x=70

\begin{gathered}\sf{x = \frac{70}{2}} \\ \end{gathered}

x=

2

70

\fbox{\sf{x = 35}}

x = 35

\sf\green{Hence \ x = 35}Hence x=35

\sf\green{y = 30}y=30

\sf\green{z = 25}z=25