Solve for s : {:[-7=s+(-13)],[s=]:}

Isolate s in first equation: Look at the first equation in the system. The first equation is -7 = s + (-13). To find the value of s, we need to isolate s on one side of the equation.Add 13 to isolate s: Add 13 to both sides of the first equation to isolate s. -7 + 13 = s + (-13) + 13 This simplifies to: 6 = sSubstitute s into second equation: Substitute the value of s into the second equation to check for consistency. The second equation is simply s =, which means it is asking for the value of s. Since we have found that s = 6 from the first equation, this is consistent with the second equation.Conclude the solution: Conclude the solution. Since the value of s = 6 satisfies both equations in the system, we have found the solution to the system.

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Math Problems

Grade 6

Multiply using the distributive property

Full solution

Q. Solve for

s

\newline

\begin{array}{l} -7=s+(-13) \\ s=\square \end{array}

Isolate $s$ in first equation: Look at the first equation in the system. $\newline$ The first equation is $-7 = s + (-13)$ . To find the value of $s$ , we need to isolate $s$ on one side of the equation.
Add $13$ to isolate s: Add $13$ to both sides of the first equation to isolate s. $\newline$ $-7 + 13 = s + (-13) + 13$ $\newline$ This simplifies to: $\newline$ $6 = s$
Substitute $s$ into second equation: Substitute the value of $s$ into the second equation to check for consistency. $\newline$ The second equation is simply $s =$ , which means it is asking for the value of $s$ . Since we have found that $s = 6$ from the first equation, this is consistent with the second equation.
Conclude the solution: Conclude the solution. $\newline$ Since the value of $s = 6$ satisfies both equations in the system, we have found the solution to the system.