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If mx + qy − nx − py = 0, p − q = 2, and
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23 Jan 2020, 10:26
23 Jan 2020, 10:26
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If \(mx + qy − nx − py = 0\), \(p − q = 2\), and \(\frac{y}{x}=- \frac{1}{3}\) , then which of the following is true?
A. \(n-m=\frac{2}{3}\)
B. \(n-m=- \frac{2}{3}\)
C. \(m+n=\frac{2}{3}\)
D. \(m+n=\frac{3}{2}\)
E. \(m+n= - \frac{3}{2}\)
A. \(n-m=\frac{2}{3}\)
B. \(n-m=- \frac{2}{3}\)
C. \(m+n=\frac{2}{3}\)
D. \(m+n=\frac{3}{2}\)
E. \(m+n= - \frac{3}{2}\)
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Re: If mx + qy − nx − py = 0, p − q = 2, and
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23 Jan 2020, 11:22
23 Jan 2020, 11:22
2
Carcass wrote:
If \(mx + qy − nx − py = 0\), \(p − q = 2\), and \(\frac{y}{x}=- \frac{1}{3}\) , then which of the following is true?
A. \(n-m=\frac{2}{3}\)
B. \(n-m=- \frac{2}{3}\)
C. \(m+n=\frac{2}{3}\)
D. \(m+n=\frac{3}{2}\)
E. \(m+n= - \frac{3}{2}\)
A. \(n-m=\frac{2}{3}\)
B. \(n-m=- \frac{2}{3}\)
C. \(m+n=\frac{2}{3}\)
D. \(m+n=\frac{3}{2}\)
E. \(m+n= - \frac{3}{2}\)
Tricky!!!
First notice that we can perform some factoring in parts
We'll start by rearranging the terms to get some like terms together.
Take: \(mx + qy − nx − py = 0\)
Rearrange to get: \(mx − nx + qy − py = 0\)
Factor in parts to get: \(x(m − n) + y(q - p) = 0\)
Now subtract \(y(q - p)\) from both sides to get: \(x(m − n) = -y(q - p)\)
Rewrite the right side of the equation as follows: \(x(m − n) = y(-q + p)\)
Simplify to get: \(x(m − n) = y(p - q)\)
Divide both sides by \(x\) to get: \(m − n = \frac{y(p - q)}{x}\)
Rewrite the right side of the equation to get: \(m − n = \frac{y}{x}(p - q)\)
Replace \(p − q\), and \(\frac{y}{x}\) to get: \(m − n = -\frac{1}{3}(2)\)
Simplify to get: \(m − n = -\frac{2}{3}\)
Multiply both sides by \(-1\) to get: \(-m + n = \frac{2}{3}\)
Finally, rewrite as follows: \(n - m = \frac{2}{3}\)
Answer: A
Cheers,
Brent
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Re: If mx + qy − nx − py = 0, p − q = 2, and
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23 Jan 2020, 11:40
23 Jan 2020, 11:40
1
Solution:
\(mx + qy − nx − py = 0\) ... (i)
\(p − q = 2\) ... (ii)
\(y/x = -1/3 => x = -3y\) ... (iii)
We observe that \(p - q\) is given; also, in (i), we can take \(y\) common and get \((p - q)\). Thus, from (i), we have:
\(mx + qy − nx − py = 0\)
\(=> mx − nx + qy − py = 0\)
\(=> x(m − n) - y(p − q) = 0\)
\(=> x(m − n) - 2y = 0\)
Using (iii), we have:
\(=> -3y(m − n) - 2y = 0\)
\(=> 3y(m − n) + 2y = 0\)
\(=> y[3(m − n) + 2] = 0\)
\(=> y = 0\) OR \(3(m - n) + 2 = 0\)
\(=> y = 0\) OR \((m - n) = -2/3\)
Taking \(y\) as non-zero (else we cannot have \(y/x = -1/3\)), we have:
\((m - n) = -2/3\)
\(=> n - m = 2/3\)
\(mx + qy − nx − py = 0\) ... (i)
\(p − q = 2\) ... (ii)
\(y/x = -1/3 => x = -3y\) ... (iii)
We observe that \(p - q\) is given; also, in (i), we can take \(y\) common and get \((p - q)\). Thus, from (i), we have:
\(mx + qy − nx − py = 0\)
\(=> mx − nx + qy − py = 0\)
\(=> x(m − n) - y(p − q) = 0\)
\(=> x(m − n) - 2y = 0\)
Using (iii), we have:
\(=> -3y(m − n) - 2y = 0\)
\(=> 3y(m − n) + 2y = 0\)
\(=> y[3(m − n) + 2] = 0\)
\(=> y = 0\) OR \(3(m - n) + 2 = 0\)
\(=> y = 0\) OR \((m - n) = -2/3\)
Taking \(y\) as non-zero (else we cannot have \(y/x = -1/3\)), we have:
\((m - n) = -2/3\)
\(=> n - m = 2/3\)
