`(3/4)x + y = 1/8, (9/4)x + 3y = 3/8` Solve the system by the method of substitution.

Tuesday, January 6, 2009

$\displaystyle{\left(\frac{{3}}{{4}}\right)}{x}+{y}=\frac{{1}}{{8}}$

$\displaystyle{\left(\frac{{9}}{{4}}\right)}{x}+{3}{y}=\frac{{3}}{{8}}$

From the first equation,

$\displaystyle{y}=\frac{{1}}{{8}}-{\left(\frac{{3}}{{4}}\right)}{x}$

Substitute the value of y obtained in the second equation,

$\displaystyle{\left(\frac{{9}}{{4}}\right)}{x}+{3}{\left(\frac{{1}}{{8}}-{\left(\frac{{3}}{{4}}\right)}{x}\right)}=\frac{{3}}{{8}}$

$\displaystyle\frac{{{9}{x}}}{{4}}+\frac{{3}}{{8}}-\frac{{{9}{x}}}{{4}}=\frac{{3}}{{8}}$

(9x)/4-(9x)/4=3/8-3/8

0=0

In this case , the given system of equations has infinitely many solutions.

If the two equations are of the form,

$\displaystyle{a}_{{1}}{x}+{b}_{{1}}{y}+{c}_{{1}}={0}$

$\displaystyle{a}_{{2}}{x}+{b}_{{2}}{y}+{c}_{{2}}={0}$

and if,

$\displaystyle\frac{{a}_{{1}}}{{a}_{{2}}}=\frac{{b}_{{1}}}{{b}_{{2}}}=\frac{{c}_{{1}}}{{c}_{{2}}}$ ,

then the the given system of equations has an infinite number of solutions.

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