Use elimination to solve each system of equations. 1/3x-y=-1 and 1/5x-2/5y=-1

Mathematics

1 answer:

Yakvenalex [24]3 years ago

4 0

Answer:

x=−9 and y=−2

Step-by-step explanation:

Multiply the first equation by 2,and multiply the second equation by -5.

2(1/3x−y=−1)

−5(1/5x−2/5y=−1)

Becomes: 2/3x−2y=−2

−x+2y=5

Add these equations to eliminate y:

−1/3 x=3

Then solve −1/3x=3 for x:

−1/3x 3

−1/3 = −1/3

(Divide both sides by (-1)/3)

x=−9

Now that we've found x let's plug it back in to solve for y.

Write down an original equation:

1/3x−y=−1

Substitute−9 for x in 1/3x−y=−1:

1/3(−9)−y=−1

−y−3=−1(Simplify both sides of the equation)

−y−3+3=−1+3(Add 3 to both sides)

−y=2

−y

−1=

−1

(Divide both sides by -1)

y=−2

You might be interested in

3^x= 3*2^x solve this equation

kompoz [17]

In the equation

$3^x = 3\cdot 2^x$

divide both sides by $2^x$ to get

$\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3$

Take the base-3/2 logarithm of both sides:

$\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}$

Alternatively, you can divide both sides by $3^x$ :

$\dfrac{3^x}{3^x} = \dfrac{3\cdot 2^x}{3^x} \\\\ \implies 1 = 3 \cdot\left(\dfrac23\right)^x \\\\ \implies \left(\dfrac23\right)^x = \dfrac13$

Then take the base-2/3 logarith of both sides to get

$\log_{2/3}\left(2/3\right)^x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x \log_{2/3}\left(\dfrac23\right) = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(3^{-1}\right) \\\\ \implies \boxed{x = -\log_{2/3}(3)}$