Solve for x. {:[-30=5(x+1)],[x=]:}

Distribute and Simplify: Solve the first equation for x. The first equation is -30 = 5(x + 1). To solve for x, we first need to distribute the 5 into the parentheses. -30 = 5(x + 1) becomes -30 = 5x + 5.Isolate Variable x: Isolate the variable x on one side of the equation. To isolate x, we need to subtract 5 from both sides of the equation. -30 - 5 = 5x + 5 - 5, which simplifies to -35 = 5x.Divide to Solve: Divide both sides of the equation by 5 to solve for x. -35 ÷ 5 = 5x ÷ 5, which simplifies to -7 = x.Check Solution: Check the solution in the second equation. The second equation is simply x =, which means x can be any value. Since we have found x to be -7 from the first equation, it satisfies the second equation.

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Math Problems

Grade 6

Multiply using the distributive property

Full solution

Q. Solve for

x

\newline

\begin{array}{l} -30=5(x+1) \\ x=\square \end{array}

Distribute and Simplify: Solve the first equation for $x$ . The first equation is $-30 = 5(x + 1)$ . To solve for $x$ , we first need to distribute the $5$ into the parentheses. $-30 = 5(x + 1)$ becomes $-30 = 5x + 5$ .
Isolate Variable $x$ : Isolate the variable $x$ on one side of the equation. $\newline$ To isolate $x$ , we need to subtract $5$ from both sides of the equation. $\newline$ $-30 - 5 = 5x + 5 - 5$ , which simplifies to $-35 = 5x$ .
Divide to Solve: Divide both sides of the equation by $5$ to solve for $x$ .\[$-35$ \div $5$ = $5$x \div $5$$, which simplifies to $$-7$ = x$.
Check Solution: Check the solution in the second equation.$\newline$The second equation is simply $x =$, which means $x$ can be any value. Since we have found $x$ to be $-7$ from the first equation, it satisfies the second equation.