Question-213939

Question Number 213939 by polymathAntunes last updated on 22/Nov/24

Commented by Frix last updated on 22/Nov/24

1. All x on one side, constants to the other side (x/4)+20=(x/3) ∣−(x/3)−20 (x/4)−(x/3)=−20 2. Common denominator and add 4×3=12 ((3x)/(12))−((4x)/(12))=−20 −(x/(12))=−20 3. Multiply by denominator and (−1) −(x/(12))=−20 ∣×12 −x=−240 ∣×(−1) x=240

$$\mathrm{1}.\:\mathrm{All}\:{x}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}, \\ $$$$\:\:\:\:\:\mathrm{constants}\:\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:\mathrm{side} \\ $$$$\:\:\:\:\:\frac{{x}}{\mathrm{4}}+\mathrm{20}=\frac{{x}}{\mathrm{3}}\:\:\:\:\:\mid−\frac{{x}}{\mathrm{3}}−\mathrm{20} \\ $$$$\:\:\:\:\:\frac{{x}}{\mathrm{4}}−\frac{{x}}{\mathrm{3}}=−\mathrm{20} \\ $$$$\mathrm{2}.\:\mathrm{Common}\:\mathrm{denominator}\:\mathrm{and}\:\mathrm{add} \\ $$$$\:\:\:\:\:\mathrm{4}×\mathrm{3}=\mathrm{12} \\ $$$$\:\:\:\:\:\frac{\mathrm{3}{x}}{\mathrm{12}}−\frac{\mathrm{4}{x}}{\mathrm{12}}=−\mathrm{20} \\ $$$$\:\:\:\:\:−\frac{{x}}{\mathrm{12}}=−\mathrm{20} \\ $$$$\mathrm{3}.\:\mathrm{Multiply}\:\mathrm{by}\:\mathrm{denominator}\:\mathrm{and}\:\left(−\mathrm{1}\right) \\ $$$$\:\:\:\:\:−\frac{{x}}{\mathrm{12}}=−\mathrm{20}\:\:\:\:\:\mid×\mathrm{12} \\ $$$$\:\:\:\:\:−{x}=−\mathrm{240}\:\:\:\:\:\mid×\left(−\mathrm{1}\right) \\ $$$$\:\:\:\:\:{x}=\mathrm{240} \\ $$

Answered by a.lgnaoui last updated on 22/Nov/24

Easy! 3x+12×20=4x x=12×20 x=240 verification : ((240)/4)+20=80=((240)/3)

$$\boldsymbol{\mathrm{Easy}}! \\ $$$$\mathrm{3x}+\mathrm{12}×\mathrm{20}=\mathrm{4x}\:\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{12}×\mathrm{20} \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{240} \\ $$$$\boldsymbol{\mathrm{v}}\mathrm{e}\boldsymbol{\mathrm{rification}}\::\:\:\frac{\mathrm{240}}{\mathrm{4}}+\mathrm{20}=\mathrm{80}=\frac{\mathrm{240}}{\mathrm{3}} \\ $$

Commented by polymathAntunes last updated on 22/Nov/24

Commented by a.lgnaoui last updated on 22/Nov/24

x((1/3)−(1/4))=20 ((x(4−3))/(12))=20 ⇒ x=20×((12)/(4−3))=240 ((ax)/b)+e=((cx)/d) ⇔ { ((x((c/d)−(a/b))=e (b#0)and d#0)),((x(((bc−ad)/(bd)))=e x=((bde)/(bc−ad)))) :} Remarque: x=((bde)/(det determinant (((c a)),((d b)))))

$$\boldsymbol{\mathrm{x}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{20} \\ $$$$\frac{\boldsymbol{\mathrm{x}}\left(\mathrm{4}−\mathrm{3}\right)}{\mathrm{12}}=\mathrm{20}\:\Rightarrow\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{20}×\frac{\mathrm{12}}{\mathrm{4}−\mathrm{3}}=\mathrm{240} \\ $$$$ \\ $$$$\frac{\boldsymbol{\mathrm{ax}}}{\boldsymbol{\mathrm{b}}}+\boldsymbol{\mathrm{e}}=\frac{\boldsymbol{\mathrm{cx}}}{\boldsymbol{\mathrm{d}}}\:\:\Leftrightarrow\:\:\begin{cases}{\boldsymbol{\mathrm{x}}\left(\frac{\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{d}}}−\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}\right)=\boldsymbol{\mathrm{e}}\:\:\:\left(\boldsymbol{\mathrm{b}}#\mathrm{0}\right)\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{d}}#\mathrm{0}}\\{\boldsymbol{\mathrm{x}}\left(\frac{\boldsymbol{\mathrm{bc}}−\boldsymbol{\mathrm{ad}}}{\boldsymbol{\mathrm{bd}}}\right)=\boldsymbol{\mathrm{e}}\:\:\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{bde}}}{\boldsymbol{\mathrm{bc}}−\boldsymbol{\mathrm{ad}}}}\end{cases} \\ $$$$\boldsymbol{{Remarque}}: \\ $$$$\boldsymbol{{x}}=\frac{\boldsymbol{{bde}}}{\boldsymbol{{det}}\begin{vmatrix}{\boldsymbol{{c}}\:\:\:\:\:\:\:\boldsymbol{{a}}}\\{\boldsymbol{{d}}\:\:\:\:\:\:\:\boldsymbol{{b}}}\end{vmatrix}} \\ $$

Commented by mr W last updated on 22/Nov/24

he has even difficulty with basic addition and multiplication operations. now you explain him with determinant of matrix. a good try! :)

$${he}\:{has}\:{even}\:{difficulty}\:{with}\:{basic} \\ $$$${addition}\:{and}\:{multiplication} \\ $$$${operations}.\:{now}\:{you}\:{explain}\:{him} \\ $$$${with}\:{determinant}\:{of}\:{matrix}.\:{a}\:{good} \\ $$$$\left.{try}!\::\right) \\ $$

Answered by luj last updated on 22/Nov/24

$$\frac{\:{x}+\mathrm{80}=}{\mathrm{4}}\:\frac{{x}}{\mathrm{3}} \\ $$$$\mathrm{3}{x}+\mathrm{240}=\mathrm{4}{x} \\ $$$$\mathrm{240}={x} \\ $$

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