How much of an alloy that is $20 \%$ copper should be mixed with 600 ounces of an alloy that is $70 \%$ copper in order to get an alloy that is $40 \%$ copper? ounces

Step 1 ：Set up the equation: \(0.20x + 0.70(600) = 0.40(x + 600)\)

Step 2 ：Simplify the equation: \(0.20x + 420 = 0.40x + 240\)

Step 3 ：Subtract \(0.20x\) from both sides: \(420 = 0.20x + 240\)

Step 4 ：Subtract \(240\) from both sides: \(180 = 0.20x\)

Step 5 ：Divide both sides by \(0.20\) to solve for \(x\): \(x = 900\)

Step 6 ：Check the result: The total amount of copper in the final mixture is \(0.40(900 + 600) = 600\) ounces. The total amount of copper from the \(20\%\) copper alloy is \(0.20(900) = 180\) ounces. The total amount of copper from the \(70\%\) copper alloy is \(0.70(600) = 420\) ounces. So, the total amount of copper is indeed \(180 + 420 = 600\) ounces, which confirms that the result is correct.

Step 7 ：\(\boxed{x = 900}\)