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2 years ago

Solve the system of equations by substitution. What is the product of x and y from the solution? 2x + 3y = 12 5x − y = 13

Mathematics

1 answer:

Darina [25.2K]2 years ago

5 0

2x + 3y = 12 (1st)

5x − y = 13 (2nd) ---> y = 5x - 13

Substitute y = 5x - 13 into (1st) equation

2x + 3(5x - 13) = 12

2x + 15x - 39 = 12

17x - 39 = 12

17x = 51

x = 3

substitute x = 7 into y = 5x - 13

y = 5x - 13

y = 5(3) - 13

y = 15 - 13

y = 2

Answer

solution x = 3 and y = 2

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AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

To find the simplified form of the given expression :

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

$=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2$ ( using the property $(ab)^m=a^m.b^m$ )

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$=2^03^{-8}$

$=\frac{1}{3^8}$ ( using the property $a^{-m}=\frac{1}{a^m}$ )

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

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