# The values of x and y for the system of linear equations: y=2x-

The values of x and y for the system of linear equations:
$$\displaystyle{y}={2}{x}-{4}$$
$$\displaystyle{y}^{{{2}}}={4}{x}$$

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Rosemary McBride
Step 1
Putting the value of y from (1) in (2)
$$\displaystyle{\left({2}{x}-{4}\right)}^{{{2}}}={4}{x}$$
$$\displaystyle{4}{x}^{{{2}}}+{16}-{16}{x}={4}{x}$$
$$\displaystyle{4}{\left({x}^{{{2}}}+{4}-{4}{x}\right)}={4}{x}$$
$$\displaystyle{\left({\left({a}\pm{b}\right)}^{{{2}}}={a}^{{{2}}}+{b}^{{{2}}}\pm{2}{a}{b}\right)}$$ $$\displaystyle{x}^{{{2}}}+{4}-{4}{x}={x}$$
$$\displaystyle{x}^{{{2}}}-{5}{x}+{4}={0}$$
$$\displaystyle{x}^{{{2}}}-{4}{x}-{x}+{4}={0}$$
$$\displaystyle{x}{\left({x}-{4}\right)}-{1}{\left({x}-{4}\right)}={0}$$
$$\displaystyle{\left({x}-{4}\right)}{\left({x}-{1}\right)}={0}$$
Either $$\displaystyle{x}-{4}={0}$$ or $$\displaystyle{x}-{1}={0}$$
Either $$\displaystyle{x}={1}$$ or $$\displaystyle{x}={4}$$
Putting the value of x in (2)
$$\displaystyle{y}^{{{2}}}={4}{\left({1}\right)}={4}$$
$$\displaystyle{y}=\pm{2}$$
$$\displaystyle{y}^{{{2}}}={4}{\left({4}\right)}={16}$$
$$\displaystyle{y}=\pm{4}$$