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Rashid [163]
2 years ago
14

Solve 8=3r-1 step by step

Mathematics
2 answers:
notsponge [240]2 years ago
8 0
Hi,

Again, we're going to be following the same format we did last time, but the sides of the equation have been swapped. Let's not worry about this, as it doesn't exactly make a difference.

Since 1 is negative, we must add it on both sides.

<em>8 + 1 = 3r - 1 + 1
</em>
<em>9 = 3r
</em>
<em />Now, we have 3 which is multiplying our variable r. The inverse of multiplication is division, so we will divide both sides by 3. This will leave us with our answer.

<em>9 / 3 = 3r / 3
</em>
<em>3 = r </em>or <em>r = 3.
</em>
Hopefully, this helps.
tatuchka [14]2 years ago
7 0
1+8=3r
9=3r
Both divided by 3
3=r
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M+2/3=1/4m-1<br><br><br>what the formula used to find the value of m?​
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<u>Answer:</u>

The value of m is \frac{-5+\sqrt{34}}{6} \text { or } \frac{-5-\sqrt{34}}{6} by using quadratic formula

<u>Solution:</u>

Given, expression is m+\frac{2}{3}=\frac{1}{4 m}-1

Now, we have to solve the above given expression.

\text { Now, } \mathrm{m}+\frac{2}{3}=\frac{1}{4 m}-1

By multiplying the equation with m, we get

\begin{array}{l}{m^{2}+\frac{2}{3} m+m=\frac{1}{4}} \\\\ {m^{2}+m\left(\frac{2}{3}+1\right)=\frac{1}{4}} \\\\ {m^{2}+\frac{5}{3} m=\frac{1}{4}}\end{array}

\begin{array}{l}{12 m^{2}+20 m=3} \\ {12 m^{2}+20 m-3=0}\end{array}

Now, let us use quadratic formula

\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Here in our problem, a = 12, b = 20, c = -3

\begin{array}{l}{m=\frac{-20 \pm \sqrt{20^{2}-4 \times 12 \times(-3)}}{2 \times 12}} \\\\ {=\frac{-20 \pm \sqrt{400+144}}{24}} \\\\ {=\frac{-20 \pm \sqrt{544}}{24}} \\\\ {=\frac{-20 \pm 4 \sqrt{34}}{24}=\frac{-5 \pm \sqrt{34}}{6}} \\\\ {=\frac{-5+\sqrt{34}}{6} \text { or } \frac{-5-\sqrt{34}}{6}}\end{array}

Hence the value of m is \frac{-5+\sqrt{34}}{6} \text { or } \frac{-5-\sqrt{34}}{6} by using quadratic formula

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