Question

a) 5z+1/3 =7
b)5x/3+3 =x+7
Solve them please
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Answer 1

${\tt{\red{\underline{\huge{Answer:}}}}}$

a) $5z+\dfrac{1}{3}=7$

$5z =\dfrac{7}{1}-\dfrac{1}{3}$

$5z =\dfrac{21 - 1}{3}$

$5z =\dfrac{20}{3}$

$z =\dfrac{20}{3×5}$

$z =\dfrac{20}{15}$

$z =\dfrac{4}{3}$

b)$\dfrac{5x}{3} + 3 = x + 7$

$\dfrac{5x}{3} - \dfrac{x}{1} = 7 - 3$

$\dfrac{5x - 3x}{3} = 4$

$\dfrac{2x}{3} = 4$

$2x = 4 \times 3$

$2x = 12$

$x = \dfrac{12}{2}$

$x = 6$

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Answer 2

ANSWER✔

$\sf\large\star 5z+ \dfrac{1}{3} = 7$

$\sf\implies \dfrac{15z+1}{3}=7$

$\sf\implies 15z+1=7 \times 3$

$\sf\implies 15z= 21-1$

$\sf\implies z= \dfrac{20}{15}$

$\sf\implies z= \cancel \dfrac{20}{15}$

$\sf\implies z= \dfrac{4}{3}$

$\large{\boxed{\bf{ \star\:\: z= \dfrac{4}{3}\:\: \star}}}$

$\sf\large\star \dfrac{5x}{3} + 3= x+7$

$\sf\implies \dfrac{5x+9}{3} = x+7$

$\sf\implies 5x+9= 3(x+7)$

$\sf\implies 5x+9=3x+21$

$\sf\implies 5x-3x= 21-9$

$\sf\implies 2x= 12$

$\sf\implies x= \dfrac{12}{2}$

$\sf\implies x= \cancel \dfrac{12}{2}$

$\sf\implies x=6$

$\large{\boxed{\bf{ \star\:\: x=6\:\: \star}}}$

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