$$ \begin{array}{lll}\begin{array}{lll} ext { (a) } 3 g=6.3\end{array} \ \begin{array}{ll} ext { (b) } 5 x=12+18\end{array} \ & ext { (c) } 2(a+2 a)=90\end{array} $$

Question

$$ \begin{array}{lll}\begin{array}{lll}	ext { (a) } 3 g=6.3\end{array} \ \begin{array}{ll}	ext { (b) } 5 x=12+18\end{array} \ & 	ext { (c) } 2(a+2 a)=90\end{array} $$

UpStudy AI · Accepted Answer

**(a)**  
We start with the equation  
$$
3g = 6.3
$$  
To solve for $$g$$, divide both sides by 3:  
$$
g = \frac{6.3}{3} = 2.1
$$

---

**(b)**  
The equation is  
$$
5x = 12 + 18
$$  
First, compute the sum on the right-hand side:  
$$
12 + 18 = 30
$$  
Now the equation becomes:  
$$
5x = 30
$$  
Divide both sides by 5 to solve for $$x$$:  
$$
x = \frac{30}{5} = 6
$$

---

**(c)**  
The equation is  
$$
2(a + 2a) = 90
$$  
Combine like terms inside the parentheses:  
$$
a + 2a = 3a
$$  
Thus, the equation becomes:  
$$
2(3a) = 90 \quad \Longrightarrow \quad 6a = 90
$$  
Divide both sides by 6 to solve for $$a$$:  
$$
a = \frac{90}{6} = 15
$$
**(a)** $$ g = 2.1 $$ **(b)** $$ x = 6 $$ **(c)** $$ a = 15 $$

\( \begin{array}{lll}\begin{array}{lll}\text { (a) } 3 g=6.3\end{array} \\ \begin{array}{ll}\text { (b) } 5 x=12+18\end{array} \\ & \text { (c) } 2(a+2 a)=90\end{array} \)

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