Solve the system of equations: 3x - y = -2, 2x^2 - y = 0

Question
Equations
asked 2020-12-22
Solve the system of equations: \(\displaystyle{3}{x}-{y}=-{2},{2}{x}^{{2}}-{y}={0}\)

Answers (1)

2020-12-23
Step 1
Given: The system of equations
3x−y=−2
\(\displaystyle{2}{x}^{{2}}−{y}={0}\)
To find: The solution of given system of equation.
Step 2
Explanation:
Let
3x−y=−2 ...(i)
\(\displaystyle{2}{x}^{{2}}−{y}={0}\) ...(ii),
From equation (i),
3x−y=−2,
\(\displaystyle\Rightarrow{y}={3}{x}+{2}\)
Substituting y=3x+2 in equation (ii),
\(\displaystyle{2}{x}^{{2}}-{y}={0}\)
\(\displaystyle\Rightarrow{2}{x}^{{2}}-{\left({3}{x}+{2}\right)}={0}\)
\(\displaystyle\Rightarrow{2}{x}^{{2}}-{3}{x}-{2}={0}\)
\(\displaystyle\Rightarrow{2}{x}^{{2}}-{4}{x}+{x}-{2}={0}\)
\(\displaystyle\Rightarrow{\left({x}-{2}\right)}{\left({2}{x}+{1}\right)}={0}\)
\(\displaystyle\Rightarrow{x}={2},-\frac{{1}}{{2}}\)
Now,
For \(\displaystyle{x}={2},{y}={3}\times{2}+{2}={8}\),
For \(\displaystyle{x}=-\frac{{1}}{{2}},{y}={3}\times{\left(-\frac{{1}}{{2}}\right)}+{2}=-\frac{{3}}{{2}}+{2}=\frac{{1}}{{2}}\)
Hence x=2, y=8 and \(\displaystyle{x}=−\frac{{1}}{{2}},{y}=\frac{{1}}{{2}}\) are solution of given system of equations.
Result: x=2, y=8 and \(\displaystyle{x}=−\frac{{1}}{{2}},{y}=\frac{{1}}{{2}}\) are solution of given system of equation.
0

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