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3 years ago

How do i solve this by solving systems of equations by elimination? i actually hella confused

Mathematics

2 answers:

Dmitriy789 [7]3 years ago

3 0

follow if you like.please

horrorfan [7]3 years ago

3 0

Your solutions are Y = 27 and x = 1/5

To solve systems of equations by elimination, you need to eliminate one of the variables so that you can solve for the other one. To do this, you either add or subtract the two equations. In this case, we need to subtract because both x coefficients are positive:

Y = 35x + 20

- Y = 60x + 15

0 = -25x + 5

25x = 5

x = 1/5

Now we substitute in this value of x to one of the equations and solve for Y:

Y = 35x + 20

Y = 35(1/5) + 20

Y = 7 + 20 = 27

I hope this helps! Let me know if you have any questions :)

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Elza [17]

Answer:

Step-by-step explanation:

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8 0

3 years ago

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Arada [10]

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8 0

3 years ago

What is the inverse equation for h(x) = 2log(x-3) ?

marshall27 [118]

Given the function:

$h(x)=2log\left(x-3\right)$

To find its inverse:

$\begin{gathered} y=2log\left(x-3\right) \\ x=2log\left(y-3\right) \end{gathered}$

solving for y:

$\frac{x}{2}=log(y-3)$ $\begin{gathered} 10^{\frac{x}{2}}=y-3 \\ \\ 10^{\frac{x}{2}}+3=y \end{gathered}$

ANSWER

$h^^{-1}(x)=10^{\frac{x}{2}}+3$

8 0

1 year ago

Is this a function, explain answer

likoan [24]

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5 0

3 years ago

Read 2 more answers

Can you solve 27,28,39,30,31,32,33,34?

kozerog [31]

Answer:

27. x^7

28. 4x^13

29. 1/f^4

30. x^5

31. 2x^14

32. (4y^3 x^2)^2

33. (2y/x)^6

Step-by-step explanation:

$27. \: \: \frac{ {x}^{ - 1} }{ {x}^{ - 8} } \\ \frac{ {x}^{8} }{ {x}^{1} } \\ {x}^{8 - 1} = {x}^{7} \\$

$28. \: \: \: \frac{ {52x}^{6} }{ {13x}^{ - 7} } \\ \frac{ {52x}^{6} {x}^{7} }{13} \\ {4x}^{6 + 7} = {4x}^{13}$

$29. \: \: {f}^{ - 3} ( {f}^{2} )( {f}^{ - 3} ) \\ {f}^{( - 3) +2 + ( - 3) } \\ {f}^{ - 4} \\ \frac{1}{ {f}^{4} } \\$

$30. \: \: \frac{ {x}^{ - 4} }{ {x}^{ - 9} } \\ \frac{ {x}^{9} }{ {x}^{4} } \\ {x}^{9 - 4} = {x}^{5}$

$31. \: \: \frac{ {24x}^{6} }{ {12x}^{ - 8} } \\ \frac{ {24x}^{6} {x}^{8} }{12} \\ {2x}^{6 + 8} = {2x}^{14} \\$

$32. \: \: \frac{ {3x}^{2} {y}^{ - 3} }{ {12x}^{6} {y}^{3} } \\ \frac{ {3x}^{2} }{ {12x}^{6} {y}^{3} {y}^{3} } \\ \frac{ {x}^{2 - 6} }{ {4y}^{3 + 3} } \\ \frac{ {x}^{ - 4} }{ {4y}^{6} } \\ \frac{1}{ {4y}^{6} {x}^{4} } = \frac{1}{({ {4y}^{3} {x}^{2} ) }^{2} }$

$33. \: \: {( {2x}^{3} {y}^{ - 3}) }^{ - 2} \\ {2x}^{3 \times - 2} {y}^{ - 3 \times - 2} \\ {2x}^{ - 6} {y}^{6} \\ \frac{ {2y}^{6} }{ {x}^{6} } \\ {( \frac{2y}{x} )}^{6} \\$

4 0

2 years ago