The length of a rectangle is 3 in. more than twice its width. If the perimeter of the rectangle is 18 in., find the width of the rectangle.
2w=l-3
2w + 2l=18
solve.
Thank you.
Can you help me with this one:
Stan invested $13,000, part at 17% and part at 2%. If the total interest at the end of the year is $1,910, how much did he invest at 17%?
Let Y = amount invested at 17%.
Then 13,000 - Y = amount invested at 2%.
Y + (13,000 - Y)*0.02 = 1910.
Solve for Y.
Your questions are likely to be ignored if you post follow questions to an earlier one. It is better to post a different question under Post a New Question rathan than to piggy back on one to which a response has been given.
Ok thank you
Now I am completely lost
You don't understand something about the answer? What? Please be specific.
Correct, I do not understand how to solve for Y
To solve this problem, we can set up an equation to represent the given information.
Let's say the width of the rectangle is "w" inches. The length of the rectangle can be expressed as "2w + 3" inches, as it is 3 inches more than twice the width.
The perimeter of a rectangle is the sum of all its sides. For this problem, we have two sides with length "w" and two sides with length "2w + 3".
The perimeter of the rectangle is given as 18 inches. Setting up the equation, we get:
2w + 2(2w + 3) = 18
Simplifying this equation, we start by distributing 2 to 2w and 3:
2w + 4w + 6 = 18
Combining like terms, we have:
6w + 6 = 18
Next, we isolate the variable by subtracting 6 from both sides:
6w = 18 - 6
6w = 12
Finally, we solve for w by dividing both sides by 6:
w = 12 / 6
w = 2
Thus, the width of the rectangle is 2 inches.