$\begin{aligned} C & =832+3.2 imes 150 \ & =832+480 \ & =1312 \end{aligned}$ The cost of 150 units is $ \$1312$. xercise 18e is partly constant and partly varies as $y$. When $y=2, x=30$, and when $y=6, x=50$. a Find the relationship between $x$ and $y$. b Find $x$ when $y=3$.

Question

$\begin{aligned} C &amp; =832+3.2 	imes 150 \ &amp; =832+480 \ &amp; =1312 \end{aligned}$ The cost of 150 units is $ \$1312$. xercise 18e is partly constant and partly varies as $y$. When $y=2, x=30$, and when $y=6, x=50$. a Find the relationship between $x$ and $y$. b Find $x$ when $y=3$.

GinnyAnswer · Answer

Express the relationship between x and y as x = a + b y . 
 Substitute the given values to form a system of equations: 30 = a + 2 b and 50 = a + 6 b . 
 Solve the system to find a = 20 and b = 5 , so the relationship is x = 20 + 5 y . 
 Substitute y = 3 into the equation to find x = 20 + 5 ( 3 ) = 35 , so the final answer is 35 ​ . 
 
 Explanation 
 
 Understanding the Problem We are given that x is partly constant and partly varies as y . This means we can express the relationship between x and y as x = a + b y , where a is the constant part and b y is the part that varies with y . We are given two sets of values for x and y : when y = 2 , x = 30 , and when y = 6 , x = 50 . Our goal is to find the values of a and b and then use them to find x when y = 3 . 
 
 Setting up the Equations Using the given information, we can set up a system of two linear equations:

When y = 2 , x = 30 : 30 = a + 2 b (Equation 1) When y = 6 , x = 50 : 50 = a + 6 b (Equation 2) 
 
 Solving for b To solve for a and b , we can use the method of elimination. Subtract Equation 1 from Equation 2: 
 
 50 − 30 = ( a + 6 b ) − ( a + 2 b ) 20 = 4 b b = 4 20 ​ = 5 
 
 Solving for a Now that we have the value of b , we can substitute it back into Equation 1 to solve for a : 
 
 30 = a + 2 ( 5 ) 30 = a + 10 a = 30 − 10 = 20 
 
 Finding the Relationship Now we know that a = 20 and b = 5 . Therefore, the relationship between x and y is x = 20 + 5 y . 
 
 Finding x when y=3 To find x when y = 3 , we substitute y = 3 into the equation:

x = 20 + 5 ( 3 ) x = 20 + 15 x = 35 
 
 Final Answer Therefore, when y = 3 , x = 35 . The relationship between x and y is x = 20 + 5 y . 
 
 Examples 
 Understanding how variables relate to each other, with one part being constant and another varying, is useful in many real-life situations. For example, consider a taxi fare. The fare often includes a fixed charge plus a variable charge based on the distance traveled. If you know the fare for two different distances, you can determine the fixed charge and the per-mile charge, and then predict the fare for any distance. This kind of analysis helps in budgeting and making informed decisions about costs.

$\begin{aligned} C & =832+3.2 \times 150 \\ & =832+480 \\ & =1312 \end{aligned}$ The cost of 150 units is $ \$1312$. xercise 18e is partly constant and partly varies as $y$. When $y=2, x=30$, and when $y=6, x=50$. a Find the relationship between $x$ and $y$. b Find $x$ when $y=3$.

Answer (1)

$\begin{aligned} C &amp; =832+3.2 \times 150 \\ &amp; =832+480 \\ &amp; =1312 \end{aligned}$ The cost of 150 units is $ \$1312$. xercise 18e is partly constant and partly varies as $y$. When $y=2, x=30$, and when $y=6, x=50$. a Find the relationship between $x$ and $y$. b Find $x$ when $y=3$.

Answer (1)

Related Questions in Mathematics

[Jawaban] Given these four points: $A(3,-10), B(-1,10), C(-8,-6), D(-10,-3)$, find the coordinates of the midpoint of line segments $A B$ and $C D$. Round decimals to the nearest tenth. midpoint of $A B(x, y)=($ , ) midpoint of $C D(x, y)=($ , )

[Jawaban] Solve $6 x+8 \leq 20$ or $5+4 x \geq 33$

[Jawaban] What are the solutions to the following system? $\begin{array}{l} -2 x^2+y=-5 \\ y=-3 x^2+5 \end{array}$ A. $(\sqrt{5},-10)$ and $(-\sqrt{5},-10)$ B. $(0,2)$ C. $(1,-2)$ D. $(\sqrt{2},-1)$ and $(-\sqrt{2},-1)$

[Jawaban] Solve the following division problem: $349 lb 6 oz \div 130$ Select one of four: A. 4 lb 3 oz B. 6 lb 7 oz C. 3 lb 9 oz D. 2 lb 11 oz

[Jawaban] Choose the improper fraction equivalent to the mixed number below. $9 \frac{4}{8}$ A. $\frac{77}{80}$ B. $\frac{76}{8}$ C. $\frac{75}{9}$ D. $\frac{75}{8}$

$\begin{aligned} C & =832+3.2 \times 150 \\ & =832+480 \\ & =1312 \end{aligned}$ The cost of 150 units is $ \$1312$. xercise 18e is partly constant and partly varies as $y$. When $y=2, x=30$, and when $y=6, x=50$. a Find the relationship between $x$ and $y$. b Find $x$ when $y=3$.