Question

4z-3/5-2z-1/2=1/5-z
solve this question​

Answer 1

Answer:

Given:-

Solve : $4z - \dfrac{3}{5} - 2z - \dfrac{1}{2} = \dfrac{1}{5} - z$

To Find:-

The value of "z".

Note:-

●》Here; we will first add/subtract the terms. After adding/subtracting, we will transpose the term to other side ( only known value ) to calculate the value of "z".

●》Transposing - It is a process in which we change the side of known to other side for finding unknown value and signs are also changed in this process. For example - Negative becomes Positive, Multiple becomes Divisional.

Solution:-

$\huge\red{4z - \dfrac{3}{5} - 2z - \dfrac{1}{2} = \dfrac{1}{5} - z}$

$\huge\red{ \ \ The \ \ value \ \ of \ \ z = ?}$

☆ According to note first point~

▪︎$4z - \dfrac{3}{5} - 2z - \dfrac{1}{2} = \dfrac{1}{5} - z$

▪︎$4z - 2z - \dfrac{3}{5} - \dfrac{1}{2} = \dfrac{1}{5} - z$

☆ L.C.M of denominators 5, 2 ( L.H.S side ) = 10

▪︎$2z - \dfrac{3}{5} × \dfrac{2}{2} - \dfrac{1}{2} × \dfrac{5}{5} = \dfrac{1}{5} - z$

▪︎$2z - \dfrac{6}{10} - \dfrac{5}{10} = \dfrac{1}{5} - z$

▪︎$2z - \dfrac{11}{10} = \dfrac{1}{5} - z$

☆ According to note second point ( Transposing "-z", $- \dfrac{11}{10}$ to other side )~

▪︎$2z + z = \dfrac{1}{5} + \dfrac{11}{10}$

☆ L.C.M of denominators 5, 10 ( R.H.S side ) = 10~

▪︎$2z + z = \dfrac{1}{5} × \dfrac{2}{2} + \dfrac{11}{10} × \dfrac{1}{1}$

▪︎$3z = \dfrac{2}{10} + \dfrac{11}{10}$

▪︎$3z = \dfrac{13}{10}$

▪︎$3 × z = \dfrac{13}{10}$

▪︎$z = \dfrac{13}{10} ÷ 3$

☆ Reciprocating "÷3"~

▪︎$z = \dfrac{13}{10} × \dfrac{1}{3}$

☆ After multiplying~

▪︎$z = \dfrac{13}{30}$

$\huge\pink{The \ \ value \ \ of \ \ z = \dfrac{13}{30}}$

Checking:-

♤ Let's check for "z" that L.H.S = R.H.S or not~

• $4z - \dfrac{3}{5} - 2z - \dfrac{1}{2} = \dfrac{1}{5} - z \implies ?$

♤ Applying "z" value~

• $4 × \dfrac{13}{30} - \dfrac{3}{5} - 2 × \dfrac{13}{30} - \dfrac{1}{2} = \dfrac{1}{5} - \dfrac{13}{30} \implies ?$

• $\dfrac{52}{30} - \dfrac{3}{5} - \dfrac{26}{30} - \dfrac{1}{2} = \dfrac{1}{5} - \dfrac{13}{30} \implies ?$

♤ L.C.M of denominators 30,5,2 ( L.H.S side ) = 30; 5, 30 ( R.H.S side ) = 30~

• $\dfrac{52}{30} × \dfrac{1}{1} - \dfrac{3}{5} × \dfrac{6}{6} - \dfrac{26}{30} × \dfrac{1}{1} - \dfrac{1}{2} × \dfrac{15}{15} = \dfrac{1}{5} × \dfrac{6}{6} - \dfrac{13}{30} × \dfrac{1}{1} \implies ?$

• $\underline\dfrac{52}{30} - \underline\dfrac{18}{30} - \underline\dfrac{26}{30} - \underline\dfrac{15}{30} = \dfrac{6}{30} - \dfrac{13}{30} \implies ?$

• $\dfrac{34}{30} - \dfrac{41}{30} = - \dfrac{7}{30} \implies ?$

• $- \dfrac{7}{30} = - \dfrac{7}{30} \implies ✔$

$\huge\green{Hence, Proved : z = \dfrac{13}{30}}$

Answer:-

Hence, the value of "z" = $\dfrac{13}{30}$ .

:)

Answer 2

Given:-

Solve : 4z - \dfrac{3}{5} - 2z - \dfrac{1}{2} = \dfrac{1}{5} - z4z−

5

3

−2z−

2

1

=

5

1

−z

To Find:-

The value of "z".

Note:-

●》Here; we will first add/subtract the terms. After adding/subtracting, we will transpose the term to other side ( only known value ) to calculate the value of "z".

●》Transposing - It is a process in which we change the side of known to other side for finding unknown value and signs are also changed in this process. For example - Negative becomes Positive, Multiple becomes Divisional.

Solution:-

\huge\red{4z - \dfrac{3}{5} - 2z - \dfrac{1}{2} = \dfrac{1}{5} - z}4z−

5

3

−2z−

2

1

=

5

1

−z

\huge\red{ \ \ The \ \ value \ \ of \ \ z = ?} The value of z=?

☆ According to note first point~

▪︎4z - \dfrac{3}{5} - 2z - \dfrac{1}{2} = \dfrac{1}{5} - z4z−

5

3

−2z−

2

1

=

5

1

−z

▪︎4z - 2z - \dfrac{3}{5} - \dfrac{1}{2} = \dfrac{1}{5} - z4z−2z−

5

3

−

2

1

=

5

1

−z

☆ L.C.M of denominators 5, 2 ( L.H.S side ) = 10

▪︎2z - \dfrac{3}{5} × \dfrac{2}{2} - \dfrac{1}{2} × \dfrac{5}{5} = \dfrac{1}{5} - z2z−

5

3

×

2

2

−

2

1

×

5

5

=

5

1

−z

▪︎2z - \dfrac{6}{10} - \dfrac{5}{10} = \dfrac{1}{5} - z2z−

10

6

−

10

5

=

5

1

−z

▪︎2z - \dfrac{11}{10} = \dfrac{1}{5} - z2z−

10

11

=

5

1

−z

☆ According to note second point ( Transposing "-z", - \dfrac{11}{10}−

10

11

to other side )~

▪︎2z + z = \dfrac{1}{5} + \dfrac{11}{10}2z+z=

5

1

+

10

11

☆ L.C.M of denominators 5, 10 ( R.H.S side ) = 10~

▪︎2z + z = \dfrac{1}{5} × \dfrac{2}{2} + \dfrac{11}{10} × \dfrac{1}{1}2z+z=

5

1

×

2

2

+

10

11

×

1

1

▪︎3z = \dfrac{2}{10} + \dfrac{11}{10}3z=

10

2

+

10

11

▪︎3z = \dfrac{13}{10}3z=

10

13

▪︎3 × z = \dfrac{13}{10}3×z=

10

13

▪︎z = \dfrac{13}{10} ÷ 3z=

10

13

÷3

☆ Reciprocating "÷3"~

▪︎z = \dfrac{13}{10} × \dfrac{1}{3}z=

10

13

×

3

1

☆ After multiplying~

▪︎z = \dfrac{13}{30}z=

30

13

\huge\pink{The \ \ value \ \ of \ \ z = \dfrac{13}{30}}The value of z=

30

13

Checking:-

♤ Let's check for "z" that L.H.S = R.H.S or not~

• 4z - \dfrac{3}{5} - 2z - \dfrac{1}{2} = \dfrac{1}{5} - z \implies ?4z−

5

3

−2z−

2

1

=

5

1

−z⟹?

♤ Applying "z" value~

• 4 × \dfrac{13}{30} - \dfrac{3}{5} - 2 × \dfrac{13}{30} - \dfrac{1}{2} = \dfrac{1}{5} - \dfrac{13}{30} \implies ?4×

30

13

−

5

3

−2×

30

13

−

2

1

=

5

1

−

30

13

⟹?

• \dfrac{52}{30} - \dfrac{3}{5} - \dfrac{26}{30} - \dfrac{1}{2} = \dfrac{1}{5} - \dfrac{13}{30} \implies ?

30

52

−

5

3

−

30

26

−

2

1

=

5

1

−

30

13

⟹?

♤ L.C.M of denominators 30,5,2 ( L.H.S side ) = 30; 5, 30 ( R.H.S side ) = 30~

• \dfrac{52}{30} × \dfrac{1}{1} - \dfrac{3}{5} × \dfrac{6}{6} - \dfrac{26}{30} × \dfrac{1}{1} - \dfrac{1}{2} × \dfrac{15}{15} = \dfrac{1}{5} × \dfrac{6}{6} - \dfrac{13}{30} × \dfrac{1}{1} \implies ?

30

52

×

1

1

−

5

3

×

6

6

−

30

26

×

1

1

−

2

1

×

15

15

=

5

1

×

6

6

−

30

13

×

1

1

⟹?

• \underline\dfrac{52}{30} - \underline\dfrac{18}{30} - \underline\dfrac{26}{30} - \underline\dfrac{15}{30} = \dfrac{6}{30} - \dfrac{13}{30} \implies ?

30

52

−

30

18

−

30

26

−

30

15

=

30

6

−

30

13

⟹?

• \dfrac{34}{30} - \dfrac{41}{30} = - \dfrac{7}{30} \implies ?

30

34

−

30

41

=−

30

7

⟹?

• - \dfrac{7}{30} = - \dfrac{7}{30} \implies ✔−

30

7

=−

30

7

⟹✔

\huge\green{Hence, Proved : z = \dfrac{13}{30}}Hence,Proved:z=

30

13

Answer:-

Hence, the value of "z" = \dfrac{13}{30}

30

13

.

:)