Linear equations in 1 variable：練習題

If \( 8 - 3k = -13 \), what is the value of \( k \)?

If \( \frac{3}{5}x - 4 = \frac{1}{2}x + 1 \), what is the value of \( x \)?

In the equation \(\frac{1}{2}(4x + 10) - a(x - 3) = 17\), \(a\) is a constant. If the equation has no solution, what is the value of \(a\)?

If \( \frac{x - 3}{4} = 5 \), what is the value of \( x \) ?

If \(\frac{3}{4}x - 2 = 10\), what is the value of \(2x\)?

If \( \frac{2}{3}x + 4 = 12 \), what is the value of \( x \)?

What value of \( m \) satisfies \( 5m + 3 = 2m + 15 \)?

Two water tanks, Tank A and Tank B, are being filled. Tank A starts with \(450\) liters of water and is being filled at a rate of \(12.5\) liters per minute. Tank B starts with \(125\) liters of water and is being filled at a rate of \(25.5\) liters per minute. After how many minutes, \(m\), will Tank B contain exactly \(150\) liters more water than Tank A?

A chemist is mixing two acid solutions. Solution A is \(20\%\) acid and Solution B is \(50\%\) acid. The chemist needs to create \(60\) liters of a new mixture that is \(40\%\) acid.

Part A: Let \(x\) be the volume, in liters, of Solution A used. Write a linear equation in terms of \(x\) that represents the total amount of acid in the final mixture.
Part B: Solve the equation to find the volume of Solution A and Solution B required for the mixture.
Part C: If the chemist accidentally used \(10\) liters more of Solution B than calculated in Part B while keeping the total volume at \(60\) liters, what is the new acid percentage of the mixture?

Consider the equation involving a constant \(k\):

\(5(2x - 3) - kx = 4x + 7\)

Part A: For what value of \(k\) will the equation have no solution?
Part B: If \(k = 2\), solve the equation for \(x\).