Solve C the following equation B-2BD-2CE-2CLC-dfrac -D-1-3B-E-L-C-dfrac -B-1-D-E-L-C-dfrac -D-1-2B-2E-2L-C-dfrac -B-1-2D-2E-2L-

The correct option is A C-B-1-x2212-2D-2E-2L-Given- B-2BD-2CE-2CL-x21D2- 2CE-2CL-B-x2212-2BD-x21D2-xA0-2C-E-L-B-x2212-2BD-x21D2-xA0-C-B-x2212-2BD-2-E-L-x21D2-xA0-C-B-1-x2212-2D-2E-2L-

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Standard VIII

Mathematics

Question

Solve for $C$ the following equation $B = 2 B D + 2 C E + 2 C L$
$C = \frac{D (1 - 3 B)}{E + L}$
$C = \frac{B (1 - D)}{E + L}$
$C = \frac{D (1 - 2 B)}{(2 E + 2 L)}$
$C = \frac{B (1 - 2 D)}{(2 E + 2 L)}$

C = \frac{B (1 - 2 D)}{(2 E + 2 L)}

C = \frac{D (1 - 3 B)}{E + L}

C = \frac{D (1 - 2 B)}{(2 E + 2 L)}

C = \frac{B (1 - D)}{E + L}

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Solution

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The correct option is A $C = \frac{B (1 - 2 D)}{(2 E + 2 L)}$
Given, $B = 2 B D + 2 C E + 2 C L$
$\Rightarrow$ $2 C E + 2 C L = B - 2 B D$
$\Rightarrow$ $2 C (E + L) = B - 2 B D$
$\Rightarrow$ $C = (B - 2 B D) / 2 (E + L)$
$\Rightarrow$ $C = B (1 - 2 D) / (2 E + 2 L)$

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The correct option is A C=B(1−2D)(2E+2L)Given, B=2BD+2CE+2CL⇒ 2CE+2CL=B−2BD⇒ 2C(E+L)=B−2BD⇒ C=(B−2BD)/2(E+L)⇒ C=B(1−2D)/(2E+2L)

Solve for $C$ the following equation $B = 2 B D + 2 C E + 2 C L$
$C = \frac{D (1 - 3 B)}{E + L}$
$C = \frac{B (1 - D)}{E + L}$
$C = \frac{D (1 - 2 B)}{(2 E + 2 L)}$
$C = \frac{B (1 - 2 D)}{(2 E + 2 L)}$

The correct option is A $C = \frac{B (1 - 2 D)}{(2 E + 2 L)}$
Given, $B = 2 B D + 2 C E + 2 C L$
$\Rightarrow$ $2 C E + 2 C L = B - 2 B D$
$\Rightarrow$ $2 C (E + L) = B - 2 B D$
$\Rightarrow$ $C = (B - 2 B D) / 2 (E + L)$
$\Rightarrow$ $C = B (1 - 2 D) / (2 E + 2 L)$