Questions and Answers
What is the solution for t in the equation d=rt?
t=d/r
What is the solution for p in the equation p=114p/y?
p=py/144
What is the solution for c in the equation R=cs/d?
c=rd/s
What is the solution for b in the equation P=a+b+c?
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What is the solution for n in the equation T=m-n?
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What is the solution for b in the equation A=a+b/2?
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What is the solution for w in the equation V=lwh?
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What is the solution for y2 in the equation m=(y2-y1)/(x2-x1)?
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What is the solution for y in the equation ax+by=c?
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What is the solution for c in the equation A=(a+b+c+d)/4?
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Study Notes
Solving for Variables in Algebra
- To isolate t in the equation ( d = rt ), rearrange to ( t = \frac{d}{r} ).
- To express p in the equation ( p = \frac{114p}{y} ), solve to get ( p = \frac{py}{144} ).
- Isolate c from the equation ( R = \frac{cs}{d} ) to find ( c = \frac{rd}{s} ).
- Rearranging ( P = a + b + c ) for b gives ( b = P - a - c ).
- To solve for n in ( T = m - n ), express it as ( n = m - T ).
- Isolating b from the formula ( A = \frac{a + b}{2} ), results in ( b = 2A - a ).
- Rearranging ( V = lwh ) to solve for w, results in ( w = \frac{V}{lh} ).
- From the equation ( m = \frac{y2 - y1}{x2 - x1} ), solving for y2 gives ( y2 = mx2 - mx1 + y1 ).
- Rearranging ( ax + by = c ) for y leads to ( y = \frac{c - ax}{b} ).
- Solving ( A = \frac{a + b + c + d}{4} ) for c rearranges to ( c = 4A - a - b - d ).
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Description
This quiz features a set of practice problems focused on solving algebraic equations for various variables. Each flashcard presents a different equation and its solution, aimed to help students reinforce their algebra skills. Perfect for Algebra 1 students looking to strengthen their problem-solving abilities.