`2x - y = 0, x - y = 7` Solve the system of linear equations and check any solutions algebraically.

Sunday, November 27, 2016

EQ1: 2x-y=0

EQ2: x-y=7

To solve this system of equations, let's apply substitution method. Let's isolate the x in the second equation.

$\displaystyle{x}-{y}={7}$

$\displaystyle{x}={7}+{y}$

Then, plug-in this to the first equation.

$\displaystyle{2}{x}-{y}={0}$

$\displaystyle{2}{\left({7}+{y}\right)}-{y}={0}$

And solve for y.

$\displaystyle{14}+{2}{y}-{y}={0}$

$\displaystyle{14}+{y}={0}$

$\displaystyle{y}=-{14}$

Now that the value of y is known, solve for x. Let's plug-in y=-14 to the second equation.

$\displaystyle{x}-{y}={7}$

$\displaystyle{x}-{\left(-{14}\right)}={7}$

$\displaystyle{x}+{14}={7}$

$\displaystyle{x}={7}-{14}$

$\displaystyle{x}=-{7}$

To check, plug-in x=-7 and y=-14 to one of the original equations. Let's use the first equation.

$\displaystyle{2}{x}-{y}={0}$

$\displaystyle{2}{\left(-{7}\right)}-{\left(-{14}\right)}={0}$

$\displaystyle-{14}+{14}={0}$

$\displaystyle{0}={0}$ (True)

Therefore, the solution is (-7,-14).

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