Réitigh (5x+b)(cx+d)=10x^2-c

$$( 5 x + b ) ( c x + d ) = 10 x ^ { 2 } - c$$

Réitigh do b.

$$\left\{\begin{matrix}b=-\frac{5cx^{2}-10x^{2}+5dx+c}{cx+d}\text{, }&d\neq -cx\\b\in \mathrm{R}\text{, }&c=\sqrt[3]{10}\times 10^{\frac{2}{3}}x^{2}\text{ and }d=-10x^{3}\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Úsáid an t-airí dáileach chun 5x+b a mhéadú faoi cx+d.

$$5cx^{2}+5xd+bcx+bd=10x^{2}-c$$

Bain 5cx^{2} ón dá thaobh.

$$5xd+bcx+bd=10x^{2}-c-5cx^{2}$$

Bain 5xd ón dá thaobh.

$$bcx+bd=10x^{2}-c-5cx^{2}-5xd$$

Comhcheangail na téarmaí ar fad ina bhfuil b.

$$\left(cx+d\right)b=10x^{2}-c-5cx^{2}-5xd$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\left(cx+d\right)b=-5cx^{2}+10x^{2}-5dx-c$$

Roinn an dá thaobh faoi cx+d.

$$\frac{\left(cx+d\right)b}{cx+d}=\frac{-5cx^{2}+10x^{2}-5dx-c}{cx+d}$$

Má roinntear é faoi cx+d cuirtear an iolrúchán faoi cx+d ar ceal.

$$b=\frac{-5cx^{2}+10x^{2}-5dx-c}{cx+d}$$