How do you solve for elimination method

1 answer:

3 0

The elimination method for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an equation.

So if you have a system: x – 6 = −6 and x + y = 8, you can add x + y to the left side of the first equation and add 8 to the right side of the equation. And since x + y = 8, you are adding the same value to each side of the first equation.

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Answer:

The answers of the questions are given below :

a) = 4096
b) = 1.25
3) = m²
4) = r⁴s³
5) = a⁸/b¹²

Step-by-step explanation:

$\large{\tt{\underline{\underline{\red{QUESTION}}}}}$

$\begin{gathered}\footnotesize\boxed{\begin{array}{c|c|c}\bf\underline{Given}&\bf\underline{Solution}&\bf{\underline{Simple\: Form}}\\\\\rule{60pt}{0.5pt} &\rule{70pt}{0.5pt}& \rule{70pt}{0.5pt}\\\\ 1.\: {4}^{6} & & \\\\ 2.\: \bigg(\dfrac{2^6}{5^3} \bigg)& &\\\\ 3. \: \Big({m}^{\frac{2}{3}}\Big)\bull \Big({m}^{\frac{4}{3}}\Big) & &\\\\4. \: \big({r}^{12} {s}^{9}\big)^{ - \frac{1}{3}} &&\\\\ 5.\bigg(\dfrac{a^4}{a^6}\bigg)^{2}& &\end{array}}\end{gathered}$

$\begin{gathered}\end{gathered}$

$\large{\tt{\underline{\underline{\red{SOLUTION}}}}}$

Question. 1

>> 4⁶

$\begin{gathered}\qquad{= 4 \times 4 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 16 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 64 \times 4 \times 4 \times 4} \\ \qquad{= 256\times 4 \times 4} \\ \qquad{= 1024 \times 4} \\ \qquad{= 4096} \end{gathered}$

Hence, the answer is 4096.

$\begin{gathered}\end{gathered}$

Question. 2

>> (2⁶/5³)^-⅓

$\begin{gathered} \qquad\implies{\bigg(\frac{2^6}{5^3}\bigg)^{ - \frac{1}{3}}}\\ \\ \qquad\implies{\bigg(\frac{64}{125}\bigg)^{ - \frac{1}{3}}}\\ \\\qquad\implies{\bigg( \frac{1}{\frac{64}{125}}\bigg)^{ \frac{1}{3}}} \\ \\ \qquad\implies{\bigg( 1 \times \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\ \\ \qquad\implies{\bigg( \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\ \\\qquad\implies{\bigg( \sqrt[3]{ \frac{125}{64}}\bigg)} \\ \\ \qquad\implies{\bigg( \dfrac{5}{4} \bigg)} \\ \\ \qquad\implies{\Big( 1.25\Big)}\end{gathered}$

Hence, the answer is 1.25.

$\begin{gathered}\end{gathered}$

Question. 3

>> (m^2/3)•(m^4/3)

$\begin{gathered} \qquad{= \Big({m}^{\frac{2}{3}}\Big) \bull \Big({m}^{ \frac{4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{2}{3} + \frac{4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{2 + 4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{6}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{2}\Big)}\end{gathered}$

Hence, the answer is m².

$\begin{gathered}\end{gathered}$

Question. 4

>> (r¹² s⁹)^⅓

$\begin{gathered} \qquad\implies{\Big( {r}^{12} \: {s}^{9}\Big)^{\frac{1}{3}}}\\\\ \qquad\implies{\Big({r}^{\frac{12}{3} } \: {s}^{\frac{9}{3}}\Big)} \\ \\ \qquad\implies{\Big({r}^{\cancel{\frac{12}{3}}} \: {s}^{\cancel{\frac{9}{3}}}\Big)} \\ \\ \qquad\implies{\Big({r}^{4} \: {s}^{3}\Big)} \end{gathered}$

Hence, the answer is r⁴s³.

$\begin{gathered}\end{gathered}$

Question. 5

>> (a⁴/b⁶)^2

$\begin{gathered} \qquad{ = \Big(\frac{a^4}{b^6}\Big)^{2}} \\ \\ \qquad{ = \Big(\frac{a^{4 \times 2}}{b^{6 \times 2}}\Big)} \\ \\ \qquad{ = \Big(\frac{a^{8}}{b^{12}}\Big)} \end{gathered}$