Question

if 1/6* (2x - 5) of 12 + 5 x 4 - 13 = 3, what is the value of x?
​

Answer 1

Answer:

We have,

If 2x−

2x

1

=13

Then, squaring both side and we get,

(2x−

2x

1

)

2

=(13)

2

4x

2

+

4x

2

1

−2×2x×

2x

1

=169

4x

2

+

4x

2

1

=169+2

4x

2

+

4x

2

1

=171

4x

2

+

4x

2

1

=171

Again, squaring both side and we get,

(4x

2

+

4x

2

1

)

2

=(171)

2

16x

4

+

16x

4

1

+2×4x

2

×

4x

2

1

=29241

16x

4

+

16x

4

1

=29241−2

16x

4

+

16x

4

1

=29239

Hence, this is the answer

Answer 2

Answer:

Required value of x is $1 \frac{1}{2}$

Step-by-step explanation:

Given,

$\frac{1}{6} \times (2x - 5) \: of \: 12 + 5 \times 4 - 13 = 3$

By BODMAS rule,we can solve this problem

By BODMAS rule,

Bracket - Of - Division - Multiplication - Addition - Subtraction

So,

$\frac{1}{6} \times (2x - 5) \: of \: 12 + 5 \times 4 - 13 = 3 \\ \frac{2x \times 12}{6} - \frac{5 \times 12}{6} + 5 \times 4 - 13 = 3 \\ 4x - 10 + 20 - 13 = 3 \\ 4x = 3 + 10 - 20 + 13 \\ 4x = 6 \\ x = \frac{6}{4} \\ x = \frac{3}{2} \\ x = 1 \frac{1}{2}$

Required value of x is $1 \frac{1}{2}$

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549

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