In the equation h/5+0.6=2 We can solve by subtracting both side with 0.6 then multiply by 5. 7 will be the answer which is correct. But if we first multiple h/5 with 5 on other side 2*5 we get h+0.6=10 after solving this we get 9.4 wich if we put back in equation is wrong but all rules followed is correct??
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The rule is to multiply both sides of the equation by a non-zero constant, not to multiply the term you like best. – Sassatelli Giulio Jun 10 '22 at 22:03
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Yeah but while subtract 0.6 from h/5+0.6 we do with one side 0.6 not including h/5 – Fredrick Jun 10 '22 at 22:05
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1What can I say... Algebra tells you how to compute $5\times\left(\frac h5+0.6\right)$ and $\left(\frac h5+0.6\right)-0.6$ – Sassatelli Giulio Jun 10 '22 at 22:10
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Ohh so u mean with multiplication there we use distributive property and with subtracting 0.6 no there's no such thing we cancel out +0.6 and -0.6 – Fredrick Jun 10 '22 at 22:16
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Can you please explain subtraction property here? – Fredrick Jun 10 '22 at 22:21
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Subtraction and sum satisfy the identity $(a+b)-c=a+(b-c)$, which is best understood as a associativity of the sum applied to $(a+b)+(-c)$. – Sassatelli Giulio Jun 11 '22 at 08:26
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you have to multiply the whole left hand side by 5, so you'd get $h+3=10$
Mark
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But while doing it with method 1 we subtract 0.6 from h/5+0.6 were we don't include h/5 – Fredrick Jun 10 '22 at 21:58
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@Fredrick Think of the equals sign as a balance. If you take $0.6$ from both sides it still balances. If you multiply all the weights on both sides by $5$ is still balances. "Including the $h$ with it" is not a useful way to think. – Ethan Bolker Jun 10 '22 at 22:04
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$4(2+3) =(42)+(43) = 8+12 = 20$ and $4(2+3) =4*5 = 20$, if you don't distribute the multiplication across terms, then it doesn't work correctly – Mark Jun 10 '22 at 22:12