Question Number 186359 by normans last updated on 03/Feb/23
Answered by HeferH last updated on 03/Feb/23
Commented by HeferH last updated on 03/Feb/23
$${i}.\:\frac{{x}\:+\:{A}}{\mathrm{2}{B}\:+\:\frac{\mathrm{3}{A}}{\mathrm{2}}}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\mathrm{3}{x}\:+\:\mathrm{3}{A}\:=\:\mathrm{4}{B}\:+\:\mathrm{3}{A} \\ $$$$\:{B}\:=\:\frac{\mathrm{3}{x}}{\mathrm{4}} \\ $$$${ii}.\:\frac{\frac{\mathrm{5}{A}}{\mathrm{2}}\:+\:\frac{\mathrm{3}{x}}{\mathrm{4}}\:+\:\mathrm{7}}{{x}\:+\:{A}\:−\:\mathrm{14}}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\:\mathrm{5}{A}\:+\:\frac{\mathrm{3}{x}}{\mathrm{2}}\:+\:\mathrm{14}\:=\:\mathrm{5}{x}\:+\:\mathrm{5}{A}\:−\mathrm{70} \\ $$$$\:\:\frac{\mathrm{3}{x}}{\mathrm{2}}\:+\:\mathrm{14}\:=\:\mathrm{5}{x}\:−\mathrm{70} \\ $$$$\:\mathrm{3}{x}\:+\:\mathrm{28}\:=\:\mathrm{10}{x}\:−\:\mathrm{140} \\ $$$$\:\mathrm{168}\:=\:\mathrm{7}{x} \\ $$$$\:{x}\:=\:\mathrm{24} \\ $$
Commented by normans last updated on 03/Feb/23
$$\boldsymbol{{very}}\:\boldsymbol{{nice}}\:\boldsymbol{{Sir}} \\ $$
Answered by mr W last updated on 05/Feb/23
Commented by mr W last updated on 05/Feb/23
$${CD}=\frac{\mathrm{1}}{\mathrm{2}}×{GH}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{7}}×{AD}=\frac{\mathrm{1}}{\mathrm{7}}×{AD} \\ $$$${BD}=\frac{\mathrm{5}}{\mathrm{7}}×{EF}=\frac{\mathrm{5}}{\mathrm{7}}×\frac{\mathrm{5}}{\mathrm{7}}×{AD}=\frac{\mathrm{25}}{\mathrm{49}}×{AD} \\ $$$${AB}={AD}−{BD}=\left(\mathrm{1}−\frac{\mathrm{25}}{\mathrm{49}}\right)×{AD}=\frac{\mathrm{24}}{\mathrm{49}}×{AD} \\ $$$$\frac{{x}}{\mathrm{7}}=\frac{{AB}}{{CD}}=\frac{\mathrm{24}}{\mathrm{49}}×\frac{\mathrm{7}}{\mathrm{1}}=\frac{\mathrm{24}}{\mathrm{7}} \\ $$$$\Rightarrow{x}=\mathrm{24}\:\checkmark \\ $$