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3 years ago

Solve the system of linear equations below.

1 answer:

6 0

y = -2x + 19
y = x + 7

-2x + 19 = x + 7
x + 2x = 19 - 7
3x = 12
x = 4

y = x + 7
y = 4 + 7
y = 11

answer
x = 4, y = 11

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

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3 years ago

Help please?

Alla [95]

$\bf \begin{array}{clclll}
3^{1001}&\qquad &4^{1002}\\
\uparrow &&\uparrow \\
a&&b
\end{array}\\\\
-----------------------------\\\\
(a+b)^2-(a-b)^2\implies (a^2+2ab+b^2)-(a^2-2ab+b^2)
\\\\\\
a^2+2ab+b^2-a^2+2ab-b^2\implies 2ab+2ab\implies 4ab
\\\\\\
now\qquad 4(3^{1001})(4^{1002})=k12^{1001}\qquad 
\begin{cases}
12^{1001}\\
(4\cdot 3)^{1001}\\
4^{1001}\cdot 3^{1001}
\end{cases}$

$\bf \\\\\\
4(3^{1001})(4^{1002})=k(4^{1001}\cdot 3^{1001})\implies \cfrac{4(3^{1001})(4^{1002})}{4^{1001}\cdot 3^{1001}}=k
\\\\\\
4\cdot \cfrac{3^{1001}}{3^{1001}}\cdot \cfrac{4^{1002}}{4^{1001}}=k\implies 4\cdot 4^{1002}4^{-1001}=k\impliedby 
\begin{array}{llll}
\textit{same base}\\
\textit{add the}\\
exponents
\end{array}
\\\\\\
4\cdot 4^{1002-1001}=k\implies 4\cdot 4^1=k\implies 16=k$

6 0

4 years ago