Answer:
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The correct answer is: [B]: " (2, 5) ".
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Given:
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-5x + y = -5 ;
-4x + 2y = 2 .
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Consider the first equation:
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-5x + y = -5 ; ↔ y + (-5x) = -5 ;
↔ y - 5x = -5 ; Add "5x" to each side of the equation; to isolate "y" on one side of the equation; and to solve in terms of "y".
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y - 5x + 5x = -5 + 5x
y = -5 + 5x ; ↔ y = 5x - 5 ;
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Now, take our second equation:
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-4x + 2y = 2 ; and plug in "(5x - 5)" for "y" ; and solve for "x" :
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-4x + 2(5x - 5) = 2 ;
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Note, 2(5x - 5) = 2(5x) - 2(5) = 10x - 10 ;
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So: -4x + 10x - 10 = 2 ;
On the left-hand side of the equation, combine the "like terms" ;
-4x +10x = 6x ; and rewrite:
6x - 10 = 2 ;
Now, add "10" to each side of the equation:
6x - 10 + 10 = 2 + 10 ;
to get:
6x = 12 ; Now, divide EACH side of the equation by "6" ; to isolate "x" on one side of the equation; and to solve for "x" ;
6x/6 = 12 / 6 ;
x = 2 ;
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Now, take our first given equation; and plug our solved value for "x" ; which is "2" ; and solve for "y" ;
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-5x + y = -5 ;
-5(2) + y = -5 ;
-10 + y = -5 ; ↔
y - 10 = -5 ;
Add "10" to each side of the equation; to isolate "y" on one side of the equation; and to solve for "y" ;
y - 10 + 10 = -5 + 10 ;
y = 5 .
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So, we have, x = 2 ; and y = 5 .
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Now, let us check our work by plugging in "2" for "x" and "5" for "y" in BOTH the original first and second equations:
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first equation:
-5x + y = -5 ;
-5(2) + 5 =? -5?
-10 + 5 =? -5 ? YES!
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second equation:
-4x + 2y = 2 ;
-4(2) + 2(5) =? 2 ?
-8 + 10 =? 2 ? Yes!
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So, the answer is:
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x = 2 , y = 5 ; or, "(2, 5)" ; which is: "Answer choice: [B] " .
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Answer:
I think you forgot the context or image of that.
Answer:
0, for q ≠ 0 and q ≠ 1
Step-by-step explanation:
Assuming q ≠ 0, you want to find the value of x such that ...
q^x = 1
This is solved using logarithms.
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x·log(q) = log(1) = 0
The zero product rule tells us this will have two solutions:
x = 0
log(q) = 0 ⇒ q = 1
If q is not 0 or 1, then its value is 1 when raised to the 0 power. If q is 1, then its value will be 1 when raised to <em>any</em> power.
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<em>Additional comment</em>
The applicable rule of logarithms is ...
log(a^b) = b·log(a)
