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Question Number 198266 by Mingma last updated on 15/Oct/23

Answered by mr W last updated on 16/Oct/23

a(−4x+3y+4z−3)+b(−2x+4y+5z−5)=10x−11y+hz−k  (−4a−2b−10)x+(3a+4b+11)y+(4a+5b−h)z−3a−5b+k=0  −4a−2b−10=0  3a+4b+11=0  ⇒a=−(9/5)  ⇒b=−(7/5)  4a+5b−h=0   ⇒h=4a+5b=((71)/5) ✓  −3a−5b+k=0   ⇒k=3a+5b=((62)/5) ✓

$${a}\left(−\mathrm{4}{x}+\mathrm{3}{y}+\mathrm{4}{z}−\mathrm{3}\right)+{b}\left(−\mathrm{2}{x}+\mathrm{4}{y}+\mathrm{5}{z}−\mathrm{5}\right)=\mathrm{10}{x}−\mathrm{11}{y}+{hz}−{k} \\ $$$$\left(−\mathrm{4}{a}−\mathrm{2}{b}−\mathrm{10}\right){x}+\left(\mathrm{3}{a}+\mathrm{4}{b}+\mathrm{11}\right){y}+\left(\mathrm{4}{a}+\mathrm{5}{b}−{h}\right){z}−\mathrm{3}{a}−\mathrm{5}{b}+{k}=\mathrm{0} \\ $$$$−\mathrm{4}{a}−\mathrm{2}{b}−\mathrm{10}=\mathrm{0} \\ $$$$\mathrm{3}{a}+\mathrm{4}{b}+\mathrm{11}=\mathrm{0} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\Rightarrow{b}=−\frac{\mathrm{7}}{\mathrm{5}} \\ $$$$\mathrm{4}{a}+\mathrm{5}{b}−{h}=\mathrm{0}\: \\ $$$$\Rightarrow{h}=\mathrm{4}{a}+\mathrm{5}{b}=\frac{\mathrm{71}}{\mathrm{5}}\:\checkmark \\ $$$$−\mathrm{3}{a}−\mathrm{5}{b}+{k}=\mathrm{0}\: \\ $$$$\Rightarrow{k}=\mathrm{3}{a}+\mathrm{5}{b}=\frac{\mathrm{62}}{\mathrm{5}}\:\checkmark \\ $$

Commented by Mingma last updated on 16/Oct/23

Nice one, sir!

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