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3 years ago

Substitution method for 7x-2y=1 & x-2y=1

1 answer:

6 0

We want to use the Substitution method to solve
7x - 2y = 1 (1)
x - 2y = 1 (2)

From equation (1), obtain
2y = 7x - 1 (3)

Substitute (3) into (2).
x - (7x - 1) = 1
-6x + 1 = 1
-6x = 0
x = 0

From (3), obtain
2y = 7*0 - 1 = -1
y = -1/2

Answer: (0, -1/2) or x=0, y= -1/2.

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3 years ago

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$\sf\dfrac{4}{7}\times\dfrac{1}{4}\rightarrow\dfrac{4\times 1}{7\times 4}\rightarrow\dfrac{4}{28}\rightarrow\dfrac{1}{7}$

$\sf\dfrac{4}{7}\times\dfrac{1}{6}\rightarrow\dfrac{4\times 1}{7\times 6}\rightarrow\dfrac{4}{42}\rightarrow\dfrac{2}{21}$

To find out which one is bigger, we need a common denominator. The LCM of 7 and 21 is 21, so convert the first fraction into a denominator of 21. 7 goes into 21 three times, multiply this to the numerator and denominator:

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Now just compare the numerators to see which one is bigger.

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3 years ago

Simplify this please

Ugo [173]

Answer:

$\frac{12q^{\frac{7}{3}}}{p^{3}}$

Step-by-step explanation:

Here are some rules you need to simplify this expression:

Distribute exponents: When you raise an exponent to another exponent, you multiply the exponents together. This includes exponents that are fractions. $(a^{x})^{n} = a^{xn}$

Negative exponent rule: When an exponent is negative, you can make it positive by making the base a fraction. When the number is apart of a bigger fraction, you can move it to the other side (top/bottom). $a^{-x} = \frac{1}{a^{x}}$ , and to help with this question: $\frac{a^{-x}b}{1} = \frac{b}{a^{x}}$ .

Multiplying exponents with same base: When exponential numbers have the same base, you can combine them by adding their exponents together. $(a^{x})(a^{y}) = a^{x+y}$

Dividing exponents with same base: When exponential numbers have the same base, you can combine them by subtracting the exponents. $\frac{a^{x}}{a^{y}} = a^{x-y}$

Fractional exponents as a radical: When a number has an exponent that is a fraction, the numerator can remain the exponent, and the denominator becomes the index (example, index here ∛ is 3). $a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = (\sqrt[n]{a})^{m}$

$\frac{(8p^{-6} q^{3})^{2/3}}{(27p^{3}q)^{-1/3}}$ Distribute exponent

$=\frac{8^{(2/3)}p^{(-6*2/3)}q^{(3*2/3)}}{27^{(-1/3)}p^{(3*-1/3)}q^{(-1/3)}}$ Simplify each exponent by multiplying

$=\frac{8^{(2/3)}p^{(-4)}q^{(2)}}{27^{(-1/3)}p^{(-1)}q^{(-1/3)}}$ Negative exponent rule

$=\frac{8^{(2/3)}q^{(2)}27^{(1/3)}p^{(1)}q^{(1/3)}}{p^{(4)}}$ Combine the like terms in the numerator with the base "q"

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(2)}q^{(1/3)}}{p^{(4)}}$ Rearranged for you to see the like terms

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(2)+(1/3)}}{p^{(4)}}$ Multiplying exponents with same base

$=\frac{8^{(2/3)}27^{(1/3)}p^{(1)}q^{(7/3)}}{p^{(4)}}$ 2 + 1/3 = 7/3

$=\frac{\sqrt[3]{8^{2}}\sqrt[3]{27}p\sqrt[3]{q^{7}}}{p^{4}}$ Fractional exponents as radical form

$=\frac{(\sqrt[3]{64})(3)(p)(q^{\frac{7}{3}})}{p^{4}}$ Simplified cubes. Wrote brackets to lessen confusion. Notice the radical of a variable can't be simplified.