Answer: B. 3x
Step-by-step explanation:
The answer is: {4, 1} .
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Explanation:
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Given the equations:
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"12x = 54 −<span> 6y ;
"-17x = -62 </span>− 6y ;
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Multiply the second equation (both sides) by "-1" ;
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-1*{-17x = -62 − 6y} ;
to get:
17x = 62 + 6y ;
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{Note: The reason we do this is that we notice the TWO "-6y" values; and by multiplying one of the entire equations by "-1" ; we can change said equation to an equation with a "(+6y)" value; and the "(+6y)" and the "(-6y)" values cancel out to "zero" ; providing an opportunity to isolate "x"; and to solve for "x".}.
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Now, rewrite the two equations:
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12x = 54 − 6y ;
17x = 62 + 6y ;
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↔ Rewrite; and then add the two together:
6y + 62 = 17x
-6y + 54 = 12x
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0 + 116 = 29x ;
↔ 29x = 116 ;
Divide EACH SIDE of the equation by "29" ; to isolate "x" on one side of the <span>equation ; and</span> to solve for "x" ;
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29x / 29 = 116 / 29 ;
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to get:
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x = 4 .
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Now that we have the value for "x" , which is "4" ; let us plug in "4" for "x" for either of the original two equations, to solve for "y". In fact, let us try substituing "4" for "x" ; for BOTH of the two original equations; to see if the value is correct.
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Our original two given equations are:
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"12x = 54 − 6y ;
"-17x = -62 − 6y ;
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Let us start with the first equation:
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" 12x = 54 − 6y " ;
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When "x = 4" ; what does "y" equal ?
Plug in "4" for "x" ; to solve for "y" ;
12(4) = 54 − 6y ;
→ 48 = 54 − 6y ;
Subtract "54" from EACH SIDE of the equation;
→ 48 − 54 = 54 − 6y − 54 ;
to get:
→ -6 = -6y ;
→ Divide EACH side of the equation by "-6" ; to isolate "y" on one side of the <span>equation ; and</span> to solve for "y" ;
→ -6/-6 = -6y / -6 ;
to get:
1 = y ; ↔ y = 1 ; So, we have: x = 4, y = 1 ; or, {4, 1}.
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Let us check to see, if the second (orginal equation) holds true when "x = 4" and "y = 1 " ;
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The second "original equation" given is:
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" -17x = -62 − 6y " ;
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→ -17(4) = ? -62 − 6(1) ?
→ -68 = ? -62 − 6 ?
→ -68 = ? -68 ? Yes!
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The answer is: {4, 1} .
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Short Answer: Tony 40 Cleo 30
Givens
T = Tony
C = Cleo
Equations
1/2 T + 1/3 C = 30 (1)
2/5 T + 1/2 C = 31 (2)
Adjustments
Multiply (1) by 3
Multiply (2) by 2
New Equations
3/2 T + C = 90 (1a)
<u>4/5 T + C = 62</u> (2a) Subtract (2a) from (1a)
(3/2 - 4/5)T = 28 the common multiple of 2 and 5 = 10.
(15/10 - 8/10)T = 28
7/10 T = 28 Multiply by 10
7T = 28*10
7T = 280 Divide by 7
T = 280/7 = 40
Put T = 40 into (1)
1/2 T + 1/3 C = 30
1/2 (40) + 1/3 C = 30
20 + 1/3 C = 30 Subtract 20 from both sides.
1/3 C = 30 - 20
1/3 C = 10 Multiply through by 3
C = 10 * 3
C = 30
Answer:
Cleo has 30 books.
Tony has 40 books.
Check
<em>Use Equation 2</em>
2/5 T + 1/2 C = 31
2/5*40 + 1/2 * 30 = ? 31
16 + 15 =?3`
31 = 31 They are equal.
We have the next two equations:
And
We need to find f(3)/g(3)
First, let's find f(3):
Then g(3):
Now, the equation f(3)/g(3) :
-27/11 is the result.
Yes, stop signs are octagon shaped. So, true.
