2 years ago

How do you solve:6r+7=13+7r

2 answers:

7 0

You need to get the r by itself on one side, with the constants (numbers without the r) on the other side.

$6r+7=13+7r\\6r-7r+7=13+7r-7r\\-r+7=13\\-r+7-7=13-7\\-r=6\\-r/-1=6/-1\\r=-6$

densk [106]2 years ago

5 0

$6r+7=13+7r \\ -6r\ \ \ \ \ -6r \\ 7=13+r \\ -13\ \ -13 \\ -6=r$

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J^2 over 2j times 2j over 3g

Serhud [2]

Answer:

Step-by-step explanation:

given $\frac{j^2}{2j}$ × $\frac{2j}{3g}$

the 2j on the denominator of the first fraction cancels with the 2j on the numerator of the second fraction, leaving the expression in simplified form as

= $\frac{j^2}{3g}$ → D