Question

x = -1
work:
1y + 3(0) = 1
1y + \underline{o} = -1 (work)
y = \underline{\quad}

Explanation:

Step1: Simplify the left - hand side

We have the equation \(1y + 3(0)=1\). First, calculate \(3(0)\), which is equal to \(0\). So the equation becomes \(1y+0 = 1\).

Step2: Solve for \(y\)

The equation \(1y + 0=1\) can be simplified to \(y = 1\) (since \(1y=y\) and adding \(0\) to \(y\) does not change its value).

Answer:

For the blank in \(1y+\underline{0}=- 1\) (note: there seems to be a typo in the original problem as the right - hand side should be \(1\) to be consistent with the first equation, assuming it's a typo and the right - hand side is \(1\)) the value is \(0\), and for \(y=\underline{1}\) (if we correct the right - hand side to \(1\)) or if we follow the wrong right - hand side of \(- 1\), there is an inconsistency. But based on the first equation \(1y + 3(0)=1\), the correct value of \(y\) is \(1\) and the blank in \(1y+\underline{0}=-1\) (assuming a typo in the right - hand side sign) should be \(0\). If we strictly follow the wrong right - hand side of \(-1\), there is an error in the problem setup. But if we take the first equation \(1y+3(0) = 1\) as correct, then:

For the blank in \(1y+\underline{0}=-1\) (assuming a sign error in the right - hand side, should be \(1\)): \(0\)

For \(y=\underline{1}\)