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2 years ago

What is the solution to the system of equations?

Mathematics

2 answers:

nataly862011 [7]2 years ago

8 0

<h2>Steps:</h2>

So with this system of equations, I will be using the elimination method. With the two equations, add them together and you will get:

$2x=6$

Since the y-variable has been eliminated, we can solve for x. With this, divide both sides by 2:

$x=3$

Now that we know the value of x, substitute it into either equation to solve for y:

$3+y=2\\y=-1\\\\3-y=4\\-y=1\\y=-1$

<h2>Answer:</h2>

<u>In (x,y) form, your answer is (3,-1).</u>

zysi [14]2 years ago

7 0

Answer:

x=3, y=-1

Step-by-step explanation:

x+y = 2

x -y =4

I will use elimination to eliminate the variable y. Add the two equations together.

x+y = 2

x -y =4

-------------

2x = 6

Divide each side by 2

2x/2 = 6/2

x = 3

But we still don't know y

x+y =2

Substitute x=3 into the equation

3+y =2

Subtract 3 from each side

3-3 +y =2-3

y =-1

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2 years ago

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2 years ago

Cuanto es (2x+2y=20) (-2x-6y=-52) con método de igualación y método de suma y resta

Nataly_w [17]

x =2 y =8

2x + 2y=20

2x=20-2y

x=10-y

Ahora len la otra ecuacion remplazamos x por (10-y)

-2(10-y)-6y=-52

-20+2y-6y=-52

2y-6y= -32 (-52+20= -32)

-4y=-32

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2*2+2*8=20

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3 years ago

How many minutes would he take to run 4 miles?

lana66690 [7]

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5 0

1 year ago

Situation:

Gala2k [10]

Answer:

5.5 days (nearest tenth)

Step-by-step explanation:

<u>Given formula:</u>

$\sf N=N_0e^{-kt}$

$\sf N_0$ = initial mass (at time t=0)
N = mass (at time t)
k = a positive constant
t = time (in days)

Given values:

$\sf N_0$ = 11 g
k = 0.125

Half-life: The <u>time</u> required for a quantity to reduce to <u>half of its initial value</u>.

To find the time it takes (in days) for the substance to reduce to half of its initial value, substitute the given values into the formula and set N to half of the initial mass, then solve for t:

$\begin{aligned}\sf N & = \sf N_0e^{-kt}\\\\\implies \sf \dfrac{11}{2} & = \sf 11e^{-0.125t}\\\\\sf \dfrac{1}{2} & = \sf e^{-0.125t}\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf \ln e^{-0.125t\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t \ln e\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t(1)\\\sf t & = \sf \dfrac{\ln \left(\dfrac{1}{2}\right)}{-0.125}\\\\\sf t & = \sf 5.545177444...\\\\\sf t & = \sf 5.5\:\:days\:\:(nearest\:tenth)\end{aligned}$

Therefore, the substance's half-life is 5.5 days (nearest tenth).

Learn more about solving exponential equations here:

brainly.com/question/28016999

5 0

1 year ago

Read 2 more answers