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3 years ago

\begin{aligned} &-5y=-5 \\\\ &7x+6y=7 \end{aligned}

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

7 0

Answer:

(5,1) is not a solution of the above system of equations.

Step-by-step explanation:

We are given a system of equations that are

- 5y = - 5 ......... (1) and

7x + 6y = 7 ........... (2)

So, solving equation (1) we get y = 1 and putting y = 1 in the equation (2) we get 7x = 7 - 6(1) = 1

⇒ $x = \frac{1}{7}$

So, it clear from the solution of the above equations i.e. ( $\frac{1}{7}, 1$ ), we can say that (5,1) is not a solution of the above system of equations. (Answer)

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Find the first three terms in the expansion , in ascending power of x , of (2+x)^6 and obtain the coefficient of x^2 in the expa

Nataly_w [17]

Answer:

The first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

coefficient of $x^{2}$ in the expansion of $(2+x - x^{2})^{6}$ = (240 - 192) = 48

Step-by-step explanation:

$(2+x)^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times x^{k} \times 2^{6 - k})$

= $6_{C_{0}} \times x^{0} \times 2^{6} + 6_{C_{1}} \times x^{1} \times 2^{5} + 6_{C_{2}} \times x^{2} \times 2^{4}$ + terms involving higher powers of x

= $64 + 192 \times x^{1} + 240 \times x^{2}$ + terms involving higher powers of x

so, the first 3 terms in the expansion of $(2 + x)^{6}$ , in ascending power of x are,

$64 , 192 \times x^{1} {\textrm{ and }}240 \times x^{2}$

Again,

$(2+x - x^{2})^{6}$

= $\sum_{k=0}^{6}(6_{C_{k}} \times (2 + x)^{k} \times (-x^{2})^{6 - k})$

Now, by inspection,

the term $x^{2}$ comes from k =5 and k = 6

for k = 5, the coefficient of $x^{2}$ is , $(-32) \times 6$ = -192

for k = 6 , the coefficient of $x^{2}$ is, $6_{C_{2}} \times 2^{4}$ = 240

so, coefficient of $x^{2}$ in the final expression = (240 - 192) = 48

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