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3 years ago

-3x-3y=3 Y=-5x-17 How do I solve this?

Mathematics

1 answer:

docker41 [41]3 years ago

4 0

We are have a system of equations with 2 variables, x and y

Equation 1: -3x-3y = 3

Equation 2: y = -5x-17

For equation 1, the y is by itself on one side of the equation, so it is really easily to substitute it into equation 2. In general, to solve for system of equations, you either do elimination or substitution. You just have to find which is easier for this situation.

By substituting y for -5x-17 in equation 1, there is only 1 variable which we can solve for.

-3x-(-5x-17) = 3

Distribute negative sign to (-5x-17) Avoid the common error of forgetting!

-3x+5x+17 = 3

Combine like terms- same variables are like terms (ex. 4x and 29x) and all constants are like terms (ex. 7 and 189 )

2x+17 = 3

Subtract 17 from both sides so all the variables are on one side and the constants on the other

2x = -14

Divide both sides by 2 to isolate the x

x = -7

We now know that x = -7 so we need to plug it into an equation to solve for y

Equation 2: y = -5x-17

y = -5 (-7) -17

y = 18

Now we know that x = -7 and y = 8

We can write it as (-7, 8)

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Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

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Answer:

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