Given:
Width of a rectangle = 
Length of the rectangle = 
To find:
The ratio of the width to the length.
Solution:
We need to find the ratio of the width to the length.

Putting the given values, we get





Therefore, the ratio of the width to the length is 1:3.
No clue sorry i tried my best
Answer:
Step-by-step explanation:
4 (x + 3) = 2x + 8 <----"four times the sum of a number and three is equal to eight more than twice the number."
Second, we want to solve the equation:
4 (x + 3) = 2x + 8
4x + 12 = 2x + 8
2x + 12 = 8 <-------subtract 2x from both sides of the equation.
2x = -4 <-------subtract -12 from both sides
x = -2 <-------divide both sides by 2 which leaves a negative answer
Third, we want to check the answer by substituting the value of -2 for x in the original equation.
4 (x + 3) = 2x + 8
4 (-2 +3) = 2(-2) + 8
-8 + 12 = -4 + 8
4 = 4
4 (x + 3) = 2x + 8 <----"four times the sum of a number and three is equal to eight more than twice the number."
Second, we want to solve the equation:
4 (x + 3) = 2x + 8
4x + 12 = 2x + 8
2x + 12 = 8 <-------subtract 2x from both sides of the equation.
2x = -4 <-------subtract -12 from both sides
x = -2 <-------divide both sides by 2 which leaves a negative answer
Third, we want to check the answer by substituting the value of -2 for x in the original equation.
4 (x + 3) = 2x + 8
4 (-2 +3) = 2(-2) + 8
-8 + 12 = -4 + 8
4 = 4
Answer: (2, 5)
Step-by-step explanation:
y = 2x + 1 y = 4x − 3
Eliminate the equal sides of each equation and combine.
2x + 1 = 4x − 3
Solve 2x + 1 = 4x − 3 for x.
Move all terms containing x to the left side of the equation.
Subtract 4x from both sides of the equation.
2x + 1 − 4x = −3
Subtract 4x from 2x.
−2x + 1 = −3
Move all terms not containing x to the right side of the equation.
Subtract 1 from both sides of the equation.
−2x = −3 − 1
Subtract 1 from −3.
−2x = −4
Divide each term by −2 and simplify.
x = 2
Evaluate y when x = 2.
Substitute 2 for x. y = 4 (2) − 3
Simplify 4 (2) − 3.
Multiply 4 by 2.
y = 8 − 3
Subtract 3 from 8.
y = 5
The solution to the system is the complete set of ordered pairs that are valid solutions.
(2, 5)
The result can be shown in multiple forms.
Point Form:
(2, 5)
Equation Form:
x = 2, y = 5