Question

for each value of v, determine whether it is a solution to -8 = 2 - 5v.

v	is it a solution?
-4	∘	∗
10	∘	∘
-9	∘	∘

Explanation:

Step1: Understand the equation

We have the equation \(-8 = 2 - 5v\). To check if a value of \(v\) is a solution, we substitute \(v\) into the equation and see if both sides are equal.

Step2: Check \(v = 0\)

Substitute \(v = 0\) into the right - hand side (RHS) of the equation: \(2-5(0)=2 - 0=2\). The left - hand side (LHS) is \(-8\). Since \(-8
eq2\), \(v = 0\) is not a solution.

Step3: Check \(v=- 4\)

Substitute \(v=-4\) into the RHS: \(2-5(-4)=2 + 20 = 22\). The LHS is \(-8\). Since \(-8
eq22\), \(v=-4\) is not a solution.

Step4: Check \(v = 10\)

Substitute \(v = 10\) into the RHS: \(2-5(10)=2-50=-48\). The LHS is \(-8\). Since \(-8
eq - 48\), \(v = 10\) is not a solution.

Step5: Check \(v=-9\)

Substitute \(v=-9\) into the RHS: \(2-5(-9)=2 + 45 = 47\). The LHS is \(-8\). Since \(-8
eq47\), \(v=-9\) is not a solution. Wait, maybe we made a mistake. Wait, let's solve the equation for \(v\) first.

Solve \(-8=2 - 5v\) for \(v\):

Subtract 2 from both sides: \(-8 - 2=2-5v - 2\), so \(-10=-5v\).

Divide both sides by \(-5\): \(v = 2\). Wait, maybe there is a typo in the given \(v\) values? Wait, if we assume that maybe one of the values was mis - written. Wait, let's re - check the substitution for each value:

Wait, if \(v = 2\) (maybe a typo in the original problem's table, maybe \(v = 2\) instead of one of the given values). But according to the given values:

For \(v = 0\): RHS \(=2-0 = 2
eq - 8\) (No)

For \(v=-4\): RHS \(=2-5\times(-4)=2 + 20=22
eq - 8\) (No)

For \(v = 10\): RHS \(=2-5\times10=2 - 50=-48
eq - 8\) (No)

For \(v=-9\): RHS \(=2-5\times(-9)=2 + 45 = 47
eq - 8\) (No)

Wait, maybe the original equation was \(-8=2 + 5v\)? Let's check. If the equation is \(-8=2 + 5v\), then:

Subtract 2: \(-10 = 5v\), \(v=-2\). Still not matching.

Wait, maybe the equation is \(-8=-2-5v\)? Let's check. Then \(-8 + 2=-5v\), \(-6=-5v\), \(v=\frac{6}{5}=1.2\). No.

Wait, perhaps the user made a typo in the equation. But based on the given values, none of the given \(v\) values (0, - 4,10, - 9) are solutions to \(-8 = 2-5v\).

But let's go back to the table. The table has columns "Is it a solution?" with "Yes" and "No" and rows for \(v = 0\), \(v=-4\), \(v = 10\), \(v=-9\).

For \(v = 0\): Substitute into \(2-5v\), we get \(2\), which is not equal to \(-8\), so "No".

For \(v=-4\): Substitute into \(2-5v\), we get \(2-5\times(-4)=2 + 20 = 22
eq - 8\), so "No".

For \(v = 10\): Substitute into \(2-5v\), we get \(2-50=-48
eq - 8\), so "No".

For \(v=-9\): Substitute into \(2-5v\), we get \(2 + 45 = 47
eq - 8\), so "No".

Answer:

For \(v = 0\): No

For \(v=-4\): No

For \(v = 10\): No

For \(v=-9\): No