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2 years ago

Solve the equation : 7(x + 6) + 7x = 9 options 2 5/14, -1 5/14, -2 5/14, 5/14

Mathematics

1 answer:

Cerrena [4.2K]2 years ago

3 0

Answer:

(i) x + 3 = 0

L.H.S. = x + 3

By putting x = 3,

L.H.S. = 3 + 3 = 6 ≠ R.H.S.

∴ No, the equation is not satisfied.

(ii) x + 3 = 0

L.H.S. = x + 3

By putting x = 0,

L.H.S. = 0 + 3 = 3 ≠ R.H.S.

∴ No, the equation is not satisfied.

(iii) x + 3 = 0

L.H.S. = x + 3

By putting x = −3,

L.H.S. = − 3 + 3 = 0 = R.H.S.

∴ Yes, the equation is satisfied.

(iv) x − 7 = 1

L.H.S. = x − 7

By putting x = 7,

L.H.S. = 7 − 7 = 0 ≠ R.H.S.

∴ No, the equation is not satisfied.

(v) x − 7 = 1

L.H.S. = x − 7

By putting x = 8,

L.H.S. = 8 − 7 = 1 = R.H.S.

∴ Yes, the equation is satisfied.

(vi) 5x = 25

L.H.S. = 5x

By putting x = 0,

L.H.S. = 5 × 0 = 0 ≠ R.H.S.

∴ No, the equation is not satisfied.

(vii) 5x = 25

L.H.S. = 5x

By putting x = 5,

L.H.S. = 5 × 5 = 25 = R.H.S.

∴ Yes, the equation is satisfied.

(viii) 5x = 25

L.H.S. = 5x

By putting x = −5,

L.H.S. = 5 × (−5) = −25 ≠ R.H.S.

∴ No, the equation is not satisfied.

(ix) = 2

L.H.S. =

By putting m = −6,

L. H. S. = ≠ R.H.S.

∴No, the equation is not satisfied.

(x) = 2

L.H.S. =

By putting m = 0,

L.H.S. = ≠ R.H.S.

∴No, the equation is not satisfied.

(xi) = 2

L.H.S. =

By putting m = 6,

L.H.S. = = R.H.S.

Step-by-step explanation:

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Answer:

$x=\dfrac{3+ \sqrt{29}}{2}, \quad \dfrac{3- \sqrt{29}}{2}$

Step-by-step explanation:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Given quadratic equation:

$x^2-3x-5=0$

Define the variables:

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Substitute the defined variables into the quadratic formula and solve for x:

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Answer:

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