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Tamiku [17]
3 years ago
8

Solve the following system.

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
4 0
<h3>The solutions are:</h3>

(0, -1)\ and\ (\frac{1}{3}, 0)

<em><u>Solution:</u></em>

<em><u>Given that,</u></em>

y = 3x^2 + 2x-1-------- eqn 1\\\\\3x-y=1 ------- eqn 2

From eqn 2,

3x - y = 1

y = 3x - 1

<em><u>Substitute the above in eqn 1</u></em>

3x -1 = 3x^2 + 2x - 1\\\\3x^2 + 2x - 3x =0\\\\3x^2 -x = 0\\\\x(3x - 1) = 0\\\\Therefore,\\\\x = 0\\\\And\\\\3x - 1 = 0\\\\3x = 1\\\\x = \frac{1}{3}

<h3>When, x = 0 </h3>

Substitute x = 0 in eqn 2

3(0) - y = 1

y = -1

<h3>When x = 1/3</h3>

Substitute x = 1/3 in eqn 2

3\frac{1}{3} - y = 1\\\\ 1 - y = 1\\\\y = 0

Thus solutions are:

(0, -1)\ and\ (\frac{1}{3}, 0)

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