If a : b = c : d, prove that :
xa + yb : xc + yd = b : d
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Given,
a : b = c : d
∴ab=cd\therefore \dfrac{a}{b} = \dfrac{c}{d}∴ba=dc
Multiplying both sides by xy\dfrac{x}{y}yx:
⇒xayb=xcyd\Rightarrow \dfrac{xa}{yb} = \dfrac{xc}{yd}⇒ybxa=ydxc
Applying componendo: ⇒xa+ybyb=xc+ydyd\Rightarrow \dfrac{xa + yb}{yb} = \dfrac{xc + yd}{yd}⇒ybxa+yb=ydxc+yd
On cross-multiplication:
⇒xa+ybxc+yd=ybyd⇒xa+ybxc+yd=bd.\Rightarrow \dfrac{xa + yb}{xc + yd} = \dfrac{yb}{yd} \\[1em] \Rightarrow \dfrac{xa + yb}{xc + yd} = \dfrac{b}{d}.⇒xc+ydxa+yb=ydyb⇒xc+ydxa+yb=db.
Hence, proved that xa + yb : xc + yd = b : d.
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If xa+b−c=yb+c−a=zc+a−b\dfrac{x}{a + b - c} = \dfrac{y}{b + c - a} = \dfrac{z}{c + a - b}a+b−cx=b+c−ay=c+a−bz = 5 and a + b + c = 7; the value of x + y + z is :
35
75\dfrac{7}{5}57
57\dfrac{5}{7}75
42
5a + 7b : 5a - 7b = 5c + 7d : 5c - 7d
If (7a + 8b)(7c - 8d) = (7a - 8b)(7c + 8d);
prove that a : b = c : d.
If x = 6aba+b\dfrac{6ab}{a + b}a+b6ab, find the value of :
x+3ax−3a+x+3bx−3b\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b}x−3ax+3a+x−3bx+3b.
