How do you solve $4 (y - \frac{2}{4}) = 9 (y + \frac{1}{3})$ $4(y-2/4)=9(y+1/3)$ ?

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Answer 1 · 2015-12-20T08:33:15

$y = - 1$ $y=-1$

$4 (y - \frac{2}{4}) = 9 (y + \frac{1}{3})$ $4(y-2/4)=9(y+1/3)$

According to BEDMAS, work on the brackets first. Reduce $\frac{2}{4}$ $2/4$ to $\frac{1}{2}$ $1/2$ .

$4 (y - \frac{1}{2}) = 9 (y + \frac{1}{3})$ $4(y-1/2)=9(y+1/3)$

Make the bracketed terms have the same denominator.

$4 (\frac{2 y}{2} - \frac{1}{2}) = 9 (\frac{3 y}{3} + \frac{1}{3})$ $4((2y)/2-1/2)=9((3y)/3+1/3)$

Subtract the bracketed terms.

$4 (\frac{2 y - 1}{2}) = 9 (\frac{3 y + 1}{3})$ $4((2y-1)/2)=9((3y+1)/3)$

Reduce the fractions by cancelling.

$color(red)cancelcolor(black)4^2((2y-1)/color(red)cancelcolor(black)2)=color(blue)cancelcolor(black)9^3((3y+1)/color(blue)cancelcolor(black)3)$

Rewrite the equation.

$2(2y-1)=3(3y+1)$

Multiply.

$4y-2=9y+3$

Isolate for $y$ by bringing all terms with $y$ to the left and all without to the right.

$4y-9y=3+2$

Solve.

$-5y=5$

$y=-1$

How do you solve 4(y-2/4)=9(y+1/3) 4(y−24)=9(y+13)4(y-2/4)=9(y+1/3) ?