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Sagot :
To solve equations with variables, we use inverse operations.
Inverse operations are operations that cancel each other out.
(plus and minus, multiply and divide)
Say we had [tex]x-3=6[/tex].
According to the addition property of equality, we could add 3 to each side of that equation and it would still be equal.
[tex]x-3+3=6+3[/tex]
This would cancel out the -3 and leave us with just [tex]x=9[/tex].
Question 1) Solve [tex]2(n-7)+3=9[/tex].
Let's just start cancelling things out. Doesn't matter what order.
[tex]2(n-7)+3=9 \\ 2(n-7)+3-3=9-3\\ 2(n-7)=6 \\\\\ \frac{2(n-7)}2=\frac{6}2\\n-7=3\\\\n-7+7=3+7\\\boxed{n=10}[/tex]
Question 2) Solve [tex]4h+7h-16=6[/tex].
You can add [tex]4h[/tex] and [tex]7h[/tex] together to get [tex]11h[/tex]. The h acts sort of like a unit in that case, like 4 feet + 7 feet = 11 feet.
[tex]11h-16=6\\11h-16+16=6+16\\11h=22\\\\\frac{11h}{11}=\frac{22}{11}\\\boxed{h=2}[/tex]
Inverse operations are operations that cancel each other out.
(plus and minus, multiply and divide)
Say we had [tex]x-3=6[/tex].
According to the addition property of equality, we could add 3 to each side of that equation and it would still be equal.
[tex]x-3+3=6+3[/tex]
This would cancel out the -3 and leave us with just [tex]x=9[/tex].
Question 1) Solve [tex]2(n-7)+3=9[/tex].
Let's just start cancelling things out. Doesn't matter what order.
[tex]2(n-7)+3=9 \\ 2(n-7)+3-3=9-3\\ 2(n-7)=6 \\\\\ \frac{2(n-7)}2=\frac{6}2\\n-7=3\\\\n-7+7=3+7\\\boxed{n=10}[/tex]
Question 2) Solve [tex]4h+7h-16=6[/tex].
You can add [tex]4h[/tex] and [tex]7h[/tex] together to get [tex]11h[/tex]. The h acts sort of like a unit in that case, like 4 feet + 7 feet = 11 feet.
[tex]11h-16=6\\11h-16+16=6+16\\11h=22\\\\\frac{11h}{11}=\frac{22}{11}\\\boxed{h=2}[/tex]
2 (n-7) +3=9
remove ( ) and distribute
2n-14+3=9
combine like terms
2n-11=9
add 11 to both sides
2n=20
n=10
4h+7h-16=6
combine like terms
11h-16=6
add 16 to both sides
11h=22
h= 2
hope this helps
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