Datrys 20^2+left(1.5tright)^2-left(12+1.5tright)^2/60t=frac{34^2+{left(1.5tright)}^2-{left(30+1.5tright)}^2}{-102t}

$$\frac{ { 20 }^{ 2 } + { \left(1.5t \right) }^{ 2 } - { \left(12+1.5t \right) }^{ 2 } }{ 60t } = \frac{ { 34 }^{ 2 } + { \left(1.5t \right) }^{ 2 } - { \left(30+1.5t \right) }^{ 2 } }{ -102t }$$

Datrys ar gyfer t

$$t = \frac{32}{7} = 4\frac{4}{7} \approx 4.571428571$$

Camau ar gyfer Datrys Hafaliad Llinol

All y newidyn t ddim fod yn hafal i 0 gan nad ydy rhannu â sero wedi’i ddiffinio. Lluoswch ddwy ochr yr hafaliad wrth 1020t, lluoswm cyffredin lleiaf 60t,-102t.

$$17\left(20^{2}+\left(1.5t\right)^{2}-\left(12+1.5t\right)^{2}\right)=-10\left(34^{2}+\left(1.5t\right)^{2}-\left(30+1.5t\right)^{2}\right)$$

Cyfrifo 20 i bŵer 2 a chael 400.

$$17\left(400+\left(1.5t\right)^{2}-\left(12+1.5t\right)^{2}\right)=-10\left(34^{2}+\left(1.5t\right)^{2}-\left(30+1.5t\right)^{2}\right)$$

Ehangu \left(1.5t\right)^{2}.

$$17\left(400+1.5^{2}t^{2}-\left(12+1.5t\right)^{2}\right)=-10\left(34^{2}+\left(1.5t\right)^{2}-\left(30+1.5t\right)^{2}\right)$$

Cyfrifo 1.5 i bŵer 2 a chael 2.25.

$$17\left(400+2.25t^{2}-\left(12+1.5t\right)^{2}\right)=-10\left(34^{2}+\left(1.5t\right)^{2}-\left(30+1.5t\right)^{2}\right)$$

Defnyddio'r theorem binomaidd \left(a+b\right)^{2}=a^{2}+2ab+b^{2} i ehangu'r \left(12+1.5t\right)^{2}.

$$17\left(400+2.25t^{2}-\left(144+36t+2.25t^{2}\right)\right)=-10\left(34^{2}+\left(1.5t\right)^{2}-\left(30+1.5t\right)^{2}\right)$$

I ddod o hyd i wrthwyneb 144+36t+2.25t^{2}, dewch o hyd i wrthwyneb pob term.

$$17\left(400+2.25t^{2}-144-36t-2.25t^{2}\right)=-10\left(34^{2}+\left(1.5t\right)^{2}-\left(30+1.5t\right)^{2}\right)$$

Tynnu 144 o 400 i gael 256.

$$17\left(256+2.25t^{2}-36t-2.25t^{2}\right)=-10\left(34^{2}+\left(1.5t\right)^{2}-\left(30+1.5t\right)^{2}\right)$$

Cyfuno 2.25t^{2} a -2.25t^{2} i gael 0.

$$17\left(256-36t\right)=-10\left(34^{2}+\left(1.5t\right)^{2}-\left(30+1.5t\right)^{2}\right)$$

Defnyddio’r briodwedd ddosbarthu i luosi 17 â 256-36t.

$$4352-612t=-10\left(34^{2}+\left(1.5t\right)^{2}-\left(30+1.5t\right)^{2}\right)$$

Cyfrifo 34 i bŵer 2 a chael 1156.

$$4352-612t=-10\left(1156+\left(1.5t\right)^{2}-\left(30+1.5t\right)^{2}\right)$$

Ehangu \left(1.5t\right)^{2}.

$$4352-612t=-10\left(1156+1.5^{2}t^{2}-\left(30+1.5t\right)^{2}\right)$$

Cyfrifo 1.5 i bŵer 2 a chael 2.25.

$$4352-612t=-10\left(1156+2.25t^{2}-\left(30+1.5t\right)^{2}\right)$$

Defnyddio'r theorem binomaidd \left(a+b\right)^{2}=a^{2}+2ab+b^{2} i ehangu'r \left(30+1.5t\right)^{2}.

$$4352-612t=-10\left(1156+2.25t^{2}-\left(900+90t+2.25t^{2}\right)\right)$$

I ddod o hyd i wrthwyneb 900+90t+2.25t^{2}, dewch o hyd i wrthwyneb pob term.

$$4352-612t=-10\left(1156+2.25t^{2}-900-90t-2.25t^{2}\right)$$

Tynnu 900 o 1156 i gael 256.

$$4352-612t=-10\left(256+2.25t^{2}-90t-2.25t^{2}\right)$$

Cyfuno 2.25t^{2} a -2.25t^{2} i gael 0.

$$4352-612t=-10\left(256-90t\right)$$

Defnyddio’r briodwedd ddosbarthu i luosi -10 â 256-90t.

$$4352-612t=-2560+900t$$

Tynnu 900t o'r ddwy ochr.

$$4352-612t-900t=-2560$$

Cyfuno -612t a -900t i gael -1512t.

$$4352-1512t=-2560$$

Tynnu 4352 o'r ddwy ochr.

$$-1512t=-2560-4352$$

Tynnu 4352 o -2560 i gael -6912.

$$-1512t=-6912$$

Rhannu’r ddwy ochr â -1512.

$$t=\frac{-6912}{-1512}$$

Lleihau'r ffracsiwn \frac{-6912}{-1512} i'r graddau lleiaf posib drwy dynnu a chanslo allan -216.

$$t=\frac{32}{7}$$