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What is the solution to the equation 41/5 =15-(2x+3)

Sagot :

mariamikayla
[tex]\frac{41}{5}=15-(2x+3) \\\\ \frac{41}{5}=15-2x-3 \\\\ \frac{41}{5}=12^{(5}-2x^{(5} \\\\ 41=60-10x \\\\ 41-60=-10x \\\\ -19=-10x \\\\ 19=10x \\\\ \boxed{x=\frac{19}{10}} [/tex]
konrad509
[tex]\dfrac{41}{5}=15-(2x+3)\\ \dfrac{41}{5}=15-2x-3\\ \dfrac{41}{5}=12-2x|\cdot5\\ 41=60-10x\\ 10x=19\\ x=\dfrac{19}{10} [/tex]
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