Lets start with one of the equations.
<span>2x+6y=18
</span>Add -6y to both sides.
<span><span><span><span>2x</span>+<span>6y</span></span>+<span>−<span>6y</span></span></span>=<span>18+<span>−<span>6y
</span></span></span></span><span><span>2x</span>=<span><span>−<span>6y</span></span>+18
</span></span>Divide both sides by 2.
<span><span><span>2x</span>2</span>=<span><span><span>−<span>6y</span></span>+18</span>2
</span></span><span>x = <span><span>−<span>3y </span></span>+ 9 <--
Now lets plug this into the other equation.
</span></span>Simplify both sides of the equation.
<span><span><span>3<span>(<span><span>−<span>3y</span></span>+9</span>)</span></span>+<span>2y</span></span>=13
</span><span>Simplify
</span><span><span><span>−<span>7y</span></span>+27</span>=13
</span>Subtract 27 from both sides.
<span><span><span><span>−<span>7y</span></span>+27</span>−27</span>=<span>13−27
</span></span><span><span>−<span>7y</span></span>=<span>−14
</span></span>Divide both sides by -7.
<span><span><span>−<span>7y/</span></span><span>−7 </span></span>= <span><span>−14/</span><span>−7
</span></span></span><span>y=2
</span>Answer:
<span>y=2</span><span>
Now that we figured out the value for y, lets plug it into either equation.
</span>2x+6(-2)=18<span>
</span>Simplify both sides of the equation.
<span><span><span>2x</span>+<span><span>(6)</span><span>(<span>−2</span>)</span></span></span>=18
</span><span>Simplify
</span><span><span><span>2x</span>−12</span>=18
</span>Add 12 to both sides.
<span><span><span><span>2x</span>−12</span>+12</span>=<span>18+12
</span></span><span><span>2x</span>=30
</span>Divide both sides by 2.
<span><span><span>2x/</span>2 </span>= <span>30/2
</span></span><span>x=<span>15</span></span>
Final answer:
y = 2
x = 15
The answer to the question is 1/64
The center is at (h,k) so we have the folowing.
Answer: $120
Step-by-step explanation:
Assume the original price is x.
He sold the scooter for $130 and this was $50 less than 1.5 times what he paid for it.
Relevant formula is therefore:
1.5x - 50 = 130
1.5x = 130 + 50
x = 180/1.5
x = $120
Answer:
We have the matrix ![A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%268%264%5Cend%7Barray%7D%5Cright%5D)
To find the eigenvalues of A we need find the zeros of the polynomial characteristic 
Then
![p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda](https://tex.z-dn.net/?f=p%28%5Clambda%29%3Ddet%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4-%5Clambda%26-4%26-4%5C%5C0%26-8-%5Clambda%26-4%5C%5C0%268%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%3D%28-4-%5Clambda%29det%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-8-%5Clambda%26-4%5C%5C8%264-%5Clambda%5Cend%7Barray%7D%5Cright%5D%20%29%5C%5C%3D%28-4-%5Clambda%29%28%28-8-%5Clambda%29%284-%5Clambda%29%2B32%29%5C%5C%3D-%5Clambda%5E3-8%5Clambda%5E2-16%5Clambda)
Now, we fin the zeros of 
.
Then, the eigenvalues of A are 
of multiplicity 1 and 
of multiplicity 2.
Let's find the eigenspaces of A. For : 
.Then, we use row operations to find the echelon form of the matrix
![A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]\rightarrow\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%268%264%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-4%26-4%5C%5C0%26-8%26-4%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We use backward substitution and we obtain
1.
2.
Therefore,
For 
: 
.Then, we use row operations to find the echelon form of the matrix
![A+4I_3=\left[\begin{array}{ccc}0&-4&-4\\0&-4&-4\\0&8&8\end{array}\right] \rightarrow\left[\begin{array}{ccc}0&-4&-4\\0&0&0\\0&0&0\end{array}\right]](https://tex.z-dn.net/?f=A%2B4I_3%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-4%26-4%5C%5C0%26-4%26-4%5C%5C0%268%268%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-4%26-4%5C%5C0%260%260%5C%5C0%260%260%5Cend%7Barray%7D%5Cright%5D)
We use backward substitution and we obtain
1.
Then,
