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Len [333]
3 years ago
13

How do you find y? ∣3y−1∣+∣2y+3∣>5

Mathematics
1 answer:
Alexandra [31]3 years ago
5 0
Let's solve your inequality step-by-step.

<span><span><span><span><span>3y</span>−1</span>+<span>2y</span></span>+3</span>>5

</span>Step 1: Simplify both sides of the inequality.<span><span><span>5y</span>+2</span>>5

</span>Step 2: Subtract 2 from both sides.<span><span><span><span>
5y</span>+2</span>−2</span>><span>5−2</span></span><span><span>
5y</span>>3</span>
Step 3: Divide both sides by 5.<span><span><span>
5y/</span>5</span>><span>3/5</span></span><span>
y><span>3/<span>5

hopefully this helps and here's a link that i used to use a lot in middle school 
https://www.symbolab.com/solver/solve-for-equation-calculator
</span></span></span>
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Please someone help me to prove this. ​
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<h3><u>Answer</u> :</h3>

We know that,

\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}

\dag\bf\:sin2A=2sinA\:cosA

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

<u>Now, Let's solve</u> !

\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}

\leadsto\:\sf\dfrac{\frac{1-cos2A}{2}-\frac{1-cos2B}{2}}{\frac{2sinA\:cosA}{2}-\frac{2sinB\:cosB}{2}}

\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}

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\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}

\leadsto\:\bf{tan(A+B)}

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