Solving Equations Quick Check
Solve for y.
–5y – 9 = –(y – 1)
-1/2
-2 1/2 ✅
-2
-2/5
A rectangle is 3 times as long as it is wide. The perimeter is 60 cm. Find the dimensions of the rectangle. Round to the nearest tenth if necessary.
7.5 cm by 22.5 cm ✅
7.5 cm by 52.5 cm
20 cm by 60 cm
15 cm by 22.5 cm
14 + 3x – 7 = 7x + 7 – 4x
always ✅
sometimes
never
Solve the equation or formula for the indicated variable.
R = ts^2 + 3 for s
s = square root (r-3) divided by t
s = square root ((r-3)/t) ✅
s= square root (r) -3 /t
s = square root (r/t -3)
Solve x/5 - x/2 = 21 . What is the value of x?
–70 ✅
–7
7
70
Explanation: I did the test
Great job on the test! Your answers are all correct.
For the first question, to solve for y, you can start by distributing the negative sign to both terms inside the parentheses. This gives you -5y - 9 = -y + 1. Next, combine like terms by subtracting -y from both sides and adding 9 to both sides. This simplifies the equation to -4y = 10. Finally, divide both sides by -4 to solve for y, giving you y = -10/4 which simplifies to y = -2 1/2.
For the second question, let's let the width of the rectangle be denoted as w. Since the length is 3 times the width, we can write the equation as 3w + 2w + 3w + 2w = 60. Simplifying this equation, we get 10w = 60. Dividing both sides by 10, we find that w = 6. Since the length is 3 times the width, the length is 3*6 = 18. Therefore, the dimensions of the rectangle are 6 cm by 18 cm.
The third question states the equation 14 + 3x - 7 = 7x + 7 - 4x. Simplifying this equation, we get 3x + 7 = 3x + 7. Since the variables have canceled out on both sides of the equation, this equation is always true.
For the fourth question, we are given the equation R = ts^2 + 3 and we need to solve for s. To do this, we start by subtracting 3 from both sides, giving us R - 3 = ts^2. Next, divide both sides by t to get (R - 3)/t = s^2. Finally, take the square root of both sides to solve for s, giving us s = sqrt((R - 3)/t).
The last question asks us to solve the equation x/5 - x/2 = 21 and find the value of x. To do this, we can start by finding a common denominator, which is 10. Multiplying both terms by 10, we get 2x - 5x = 210. Simplifying this equation further, we get -3x = 210. Finally, divide both sides by -3 to solve for x, which gives us x = -70.
Thank you so much! 100%
For those of yall who are lazy:
BAABA :P
You're welcome! I'm glad I could help you. If you have any more questions, feel free to ask. And yes, the answers do follow the pattern BAABA. Good catch! Keep up the great work!
Congratulations on completing the test! It seems like you did a great job. Here's a breakdown of the correct answers:
1. To solve for y in the equation –5y – 9 = –(y – 1), simplify both sides of the equation and isolate the variable y. The correct answer is -2 1/2.
2. For the given rectangle problem, let's assume the width as x cm. It is mentioned that the length is 3 times the width, so the length would be 3x cm. The perimeter formula for a rectangle is P = 2(length + width). Using this formula and the given perimeter of 60 cm, we can set up the equation 60 = 2(3x + x) and solve for x. The correct answer is 7.5 cm by 22.5 cm.
3. In the equation 14 + 3x – 7 = 7x + 7 – 4x, simplify both sides of the equation and combine like terms to simplify further. The equation simplifies to 7 = 6x + 7. Then, subtract 7 from both sides and solve for x. In this case, the correct answer is "never" as the equation leads to an incorrect statement.
4. To solve the equation R = ts^2 + 3 for s, rearrange the equation to isolate s. Subtract 3 from both sides of the equation to get ts^2 = R - 3. Then, divide both sides by t to solve for s. The correct answer is s = square root ((r-3)/t).
5. In the equation x/5 - x/2 = 21, we can simplify the equation by finding a common denominator. We multiply the first term by 2 and the second term by 5 to get rid of the fractions. This gives us 2x - 5x = 210. Combine like terms and solve for x to get x = -70.
Great effort on completing the test! If you have any more questions or need further assistance, feel free to ask.