Answer :
Let's verify if [tex]\( x = 15 \)[/tex] is a solution to the equation [tex]\( \frac{2x}{5} = 4 \)[/tex].
1. Substitute [tex]\( x = 15 \)[/tex] into the equation:
[tex]\[ \frac{2 \cdot 15}{5} \][/tex]
2. Perform the multiplication in the numerator:
[tex]\[ 2 \cdot 15 = 30 \][/tex]
3. Divide the result by 5:
[tex]\[ \frac{30}{5} = 6 \][/tex]
Thus, the left side of the equation becomes 6.
4. Compare this result to the right side of the original equation:
The right side of the equation is given as 4.
5. Conclusion:
Since [tex]\( 6 \)[/tex] (left side) does not equal [tex]\( 4 \)[/tex] (right side), the equation [tex]\( \frac{2x}{5} = 4 \)[/tex] is NOT satisfied when [tex]\( x = 15 \)[/tex].
Therefore, [tex]\( x = 15 \)[/tex] is not a solution to the equation [tex]\( \frac{2x}{5} = 4 \)[/tex].
1. Substitute [tex]\( x = 15 \)[/tex] into the equation:
[tex]\[ \frac{2 \cdot 15}{5} \][/tex]
2. Perform the multiplication in the numerator:
[tex]\[ 2 \cdot 15 = 30 \][/tex]
3. Divide the result by 5:
[tex]\[ \frac{30}{5} = 6 \][/tex]
Thus, the left side of the equation becomes 6.
4. Compare this result to the right side of the original equation:
The right side of the equation is given as 4.
5. Conclusion:
Since [tex]\( 6 \)[/tex] (left side) does not equal [tex]\( 4 \)[/tex] (right side), the equation [tex]\( \frac{2x}{5} = 4 \)[/tex] is NOT satisfied when [tex]\( x = 15 \)[/tex].
Therefore, [tex]\( x = 15 \)[/tex] is not a solution to the equation [tex]\( \frac{2x}{5} = 4 \)[/tex].
