For each value of u , determine whether it is a solution to -54=6(u-2) . multirow(2)(}{ u ) & multicolumn(2)(|c|)( Is it a solution? ) cline ( 2 - 3 ) & Yes & No -4 & bigcirc & bigcirc -7 & bigcirc & bigcirc 1 & bigcirc & bigcirc 5 & bigcirc & bigcirc

## Step 1: Solve the equation for \( u \)<br />### The given equation is \( -54 = 6(u - 2) \). To find \( u \), we first isolate \( u \) by performing the following steps:<br />\[<br />-54 = 6(u - 2)<br />\]<br />Divide both sides by 6:<br />\[<br />-9 = u - 2<br />\]<br />Add 2 to both sides:<br />\[<br />u = -7<br />\]<br /><br />## Step 2: Verify each value of \( u \)<br />### We need to check if each given value of \( u \) (-4, -7, 1, 5) satisfies the equation \( -54 = 6(u - 2) \).<br /><br />### For \( u = -4 \):<br />Substitute \( u = -4 \) into the equation:<br />\[<br />-54 = 6(-4 - 2)<br />\]<br />\[<br />-54 = 6(-6)<br />\]<br />\[<br />-54 = -36 \quad (\text{False})<br />\]<br />So, \( u = -4 \) is not a solution.<br /><br />### For \( u = -7 \):<br />Substitute \( u = -7 \) into the equation:<br />\[<br />-54 = 6(-7 - 2)<br />\]<br />\[<br />-54 = 6(-9)<br />\]<br />\[<br />-54 = -54 \quad (\text{True})<br />\]<br />So, \( u = -7 \) is a solution.<br /><br />### For \( u = 1 \):<br />Substitute \( u = 1 \) into the equation:<br />\[<br />-54 = 6(1 - 2)<br />\]<br />\[<br />-54 = 6(-1)<br />\]<br />\[<br />-54 = -6 \quad (\text{False})<br />\]<br />So, \( u = 1 \) is not a solution.<br /><br />### For \( u = 5 \):<br />Substitute \( u = 5 \) into the equation:<br />\[<br />-54 = 6(5 - 2)<br />\]<br />\[<br />-54 = 6(3)<br />\]<br />\[<br />-54 = 18 \quad (\text{False})<br />\]<br />So, \( u = 5 \) is not a solution.