Question

name date period
practice
for exercise 1, use the balanced scale shown.

1. the scale shown is balanced. what is the weight of a diamond?

for exercises 2 and 3, determine the value of ( x ) that makes the equation true.

1. ( 3x+\frac{4}{5}=2x - 11 ) 3. ( -0.4x+9.2 = 1.6x-4.8 )

for exercises 4 - 6, answer the questions.

1. for what value(s) of ( x ) are the expressions ( 2x + 18.2 ) and ( 2.8x+1.48 ) equal?
2. what does it mean for a number to be a solution of a linear equation?
3. one person went for a run and ran at a pace of 6 miles per hour. at the same time, another person started at the 1 - mile marker and ran at a pace of 4 miles per hour. how long did they run when the first person met the second person? how far did each of the runners run?

Explanation:

Step1: Set up the equation for the balanced scale

Let the weight of a diamond be \(x\). The left - hand side of the scale has \(2x + 7.25\) and the right - hand side has \(4x+\frac{7}{3}\). Since the scale is balanced, \(2x + 7.25=4x+\frac{7}{3}\).

Step2: Rearrange the terms to solve for \(x\)

First, move the \(x\) terms to one side: \(7.25-\frac{7}{3}=4x - 2x\).
We know that \(7.25=\frac{725}{100}=\frac{29}{4}\). Then \(\frac{29}{4}-\frac{7}{3}=2x\).
Find a common denominator, which is \(12\). So \(\frac{29\times3}{4\times3}-\frac{7\times4}{3\times4}=2x\), \(\frac{87}{12}-\frac{28}{12}=2x\), \(\frac{87 - 28}{12}=2x\), \(\frac{59}{12}=2x\).

Step3: Solve for \(x\)

Divide both sides by \(2\): \(x=\frac{59}{12}\div2=\frac{59}{12}\times\frac{1}{2}=\frac{59}{24}\approx2.46\).

Step4: Solve \(3x+\frac{4}{5}=2x - 11\)

Subtract \(2x\) from both sides: \(3x-2x+\frac{4}{5}=2x-2x - 11\), \(x+\frac{4}{5}=-11\).
Subtract \(\frac{4}{5}\) from both sides: \(x=-11-\frac{4}{5}=-\frac{55 + 4}{5}=-\frac{59}{5}=-11.8\).

Step5: Solve \(-0.4x + 9.2=1.6x-4.8\)

Add \(0.4x\) to both sides: \(-0.4x+0.4x + 9.2=1.6x+0.4x-4.8\), \(9.2 = 2x-4.8\).
Add \(4.8\) to both sides: \(9.2 + 4.8=2x\), \(14 = 2x\).
Divide both sides by \(2\): \(x = 7\).

Step6: Solve \(2x + 18.2=2.8x+1.48\)

Subtract \(2x\) from both sides: \(2x-2x + 18.2=2.8x-2x+1.48\), \(18.2=0.8x + 1.48\).
Subtract \(1.48\) from both sides: \(18.2-1.48=0.8x\), \(16.72 = 0.8x\).
Divide both sides by \(0.8\): \(x=\frac{16.72}{0.8}=20.9\).

Step7: Answer for the running problem

Let \(t\) be the time (in hours) they run.
The distance the first runner runs is \(d_1 = 6t\) and the distance the second runner runs is \(d_2=4t + 1\) (since the second runner starts at the 1 - mile marker).
Set \(d_1=d_2\): \(6t=4t + 1\).
Subtract \(4t\) from both sides: \(6t-4t=4t-4t + 1\), \(2t=1\), \(t = 0.5\) hours.
The first runner runs \(d_1=6\times0.5 = 3\) miles.
The second runner runs \(d_2=4\times0.5+1=2 + 1=3\) miles.

Answer:

1. The weight of a diamond is \(\frac{59}{24}\approx2.46\).
2. \(x=-11.8\).
3. \(x = 7\).
4. \(x = 20.9\).
5. A number is a solution of a linear equation if when we substitute that number for the variable in the equation, the left - hand side of the equation is equal to the right - hand side of the equation.
6. They ran for \(0.5\) hours. Each runner ran \(3\) miles.