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navik [9.2K]
3 years ago
6

How many solutions does the system have 6 x + 2= y 3y-18x=12

Mathematics
1 answer:
shepuryov [24]3 years ago
3 0

Step-by-step explanation:

6x + 2 = y --------1

3y - 18x = 12 -----2

From 1

substitute y = 6x +2 in equation 2

3(6x + 2) - 18x = 12

18x + 6 - 18x = 12

Therefore, no solution!

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Prove that the sum of the squares of any two odd numbers is always even
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Answer:

proof below

Step-by-step explanation:

Remember that a number is even if it is expressed so n = 2k. It is odd if it is in the form 2k + 1 (k is just an integer)

Let's say we have to odd numbers, 2a + 1, and 2b + 1. We are after the sum of their squares, so we have (2a + 1)^2 + (2b + 1)^2. Now let's expand this;

(2a + 1)^2 + (2b + 1)^2 = 4a^2 + 4a + 4b + 4b^2 + 4b + 2

= 2(2a^2 + 2a + 2b^2 + 2b + 1)

Now the sum in the parenthesis, 2a^2 + 2a + 2b^2 + 2b + 1, is just another integer, which we can pose as k. Remember that 2 times any random integer, either odd or even, is always even. Therefore the sum of the squares of any two odd numbers is always even.

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3 years ago
Write a system of two equations in two variables to solve the problem. when fully extended, a ladder is 28 feet in length. if th
Alisiya [41]
Let the base of the ladder be x ft and the extension be y ft.

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3 0
3 years ago
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Solve the 3 × 3 system shown below. Enter the values of x, y, and z. X + 2y – z = –3 (1) 2x – y + z = 5 (2) x – y + z = 4 (3)
Svet_ta [14]
<h2>Answer:</h2>

\boxed{x=1, \y=-1, \ z=2}

<h2>Step-by-step explanation:</h2>

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\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}

Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:

\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}

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\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}

Step 4: solve for z, then for y, then for x:

\frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}

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x+2(-1)-2=-3 \\ \\ x-2-2=-3 \\ \\ \boxed{x=1}

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