4x+7=1+2(2x+3) linear equations

Mathematics

1 answer:

Komok [63]1 year ago

6 0

<h2>Key Ideas</h2>

Solving algebraic equations
Linear equations

<h2>Solving the Question</h2>

$4x+7=1+2(2x+3)$

⇒ Our goal is to solve for <em>x</em>. We can do this by moving everything else to the other side of the equation.

⇒ Open up the parentheses:

$4x+7=1+4x+6$

⇒ Simplify:

$4x+7=4x+7$

Because both sides are equal, we end up with <em>x = x</em>.

In this case, we can say that all values of <em>x</em> are true.

<h2>Answer</h2>

All values of <em>x</em> are true.

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<h3> 1) $x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2$ </h3><h3> 2) $x_1=-\dfrac13\,,\quad x_2=-3$ </h3>

Step-by-step explanation:

$x^2 - 7x + 9 = 0\quad\implies\quad a=1\,,\ b = -7\,,\ c=9\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-(-7)\pm\sqrt{(-7)^2-4\cdot1\cdot9}}{2\cdot1}=\dfrac{7\pm\sqrt{49-36}}2\\\\x_1=\dfrac{7+\sqrt{13}}2\,,\quad x_2=\dfrac{7-\sqrt{13}}2$

<h3>2)</h3><h3> $3x^2 + 10x=-3\\\\3x^2+10x+3=0\quad\implies\quad a=3\,,\ b =10\,,\ c=3\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-10\pm\sqrt{10^2-4\cdot3\cdot3}}{2\cdot3}= \dfrac{-10\pm\sqrt{100-36}}6\\\\x_1=\dfrac{-10+\sqrt{64}}6=\dfrac{-10+8}6=-\dfrac13\,,\qquad x_2=\dfrac{-10-8}6=-3$ </h3>