If k−2k+3=49\dfrac{k-2}{k+3} = \dfrac{4}{9}k+3k−2=94, then the value of k is
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We have:
=k−2k+3=49⇒9(k−2)=4(k+3)[By cross multiplication]⇒9k−18=4k+12⇒9k−4k=12+18[Transposing -18 to RHS and +4k to LHS]⇒5k=30⇒k=305⇒k=6\phantom{=} \dfrac{k - 2}{k + 3} = \dfrac{4}{9} \\[1em] \Rightarrow 9(k - 2) = 4(k + 3) \quad \text{[By cross multiplication]} \\[1em] \Rightarrow 9k - 18 = 4k + 12 \\[1em] \Rightarrow 9k - 4k = 12 + 18 \quad \text{[Transposing -18 to RHS and +4k to LHS]} \\[1em] \Rightarrow 5k = 30 \\[1em] \Rightarrow k = \dfrac{30}{5} \\[1em] \Rightarrow k = 6=k+3k−2=94⇒9(k−2)=4(k+3)[By cross multiplication]⇒9k−18=4k+12⇒9k−4k=12+18[Transposing -18 to RHS and +4k to LHS]⇒5k=30⇒k=530⇒k=6
Hence, option 2 is the correct option.
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If 2x+35=25x+32x + \dfrac{3}{5} = \dfrac{2}{5}x + 32x+53=52x+3, then the value of x is
1
32\dfrac{3}{2}23
2
52\dfrac{5}{2}25
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A number is 6 more than another. If their sum is 32, then the smaller number is
