IDNLearn.com is committed to providing high-quality answers to your questions. Ask your questions and receive reliable, detailed answers from our dedicated community of experts.
Sagot :
To solve for the value of [tex]\( x \)[/tex] in the equation [tex]\( \frac{1}{5}x - \frac{2}{3}y = 30 \)[/tex] when [tex]\( y = 15 \)[/tex], follow these steps:
1. Substitute [tex]\( y = 15 \)[/tex] into the equation:
[tex]\[ \frac{1}{5}x - \frac{2}{3}(15) = 30 \][/tex]
2. Simplify the term involving [tex]\( y \)[/tex]:
[tex]\[ \frac{2}{3} \times 15 = \frac{2 \times 15}{3} = \frac{30}{3} = 10 \][/tex]
So the equation becomes:
[tex]\[ \frac{1}{5}x - 10 = 30 \][/tex]
3. Isolate the term involving [tex]\( x \)[/tex] by adding 10 to both sides of the equation:
[tex]\[ \frac{1}{5}x - 10 + 10 = 30 + 10 \][/tex]
This simplifies to:
[tex]\[ \frac{1}{5}x = 40 \][/tex]
4. Solve for [tex]\( x \)[/tex] by multiplying both sides of the equation by 5:
[tex]\[ x = 40 \times 5 \][/tex]
Therefore:
[tex]\[ x = 200 \][/tex]
So, the value of [tex]\( x \)[/tex] is [tex]\( \boxed{200} \)[/tex].
1. Substitute [tex]\( y = 15 \)[/tex] into the equation:
[tex]\[ \frac{1}{5}x - \frac{2}{3}(15) = 30 \][/tex]
2. Simplify the term involving [tex]\( y \)[/tex]:
[tex]\[ \frac{2}{3} \times 15 = \frac{2 \times 15}{3} = \frac{30}{3} = 10 \][/tex]
So the equation becomes:
[tex]\[ \frac{1}{5}x - 10 = 30 \][/tex]
3. Isolate the term involving [tex]\( x \)[/tex] by adding 10 to both sides of the equation:
[tex]\[ \frac{1}{5}x - 10 + 10 = 30 + 10 \][/tex]
This simplifies to:
[tex]\[ \frac{1}{5}x = 40 \][/tex]
4. Solve for [tex]\( x \)[/tex] by multiplying both sides of the equation by 5:
[tex]\[ x = 40 \times 5 \][/tex]
Therefore:
[tex]\[ x = 200 \][/tex]
So, the value of [tex]\( x \)[/tex] is [tex]\( \boxed{200} \)[/tex].
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. IDNLearn.com has the solutions you’re looking for. Thanks for visiting, and see you next time for more reliable information.