Solve (13)/(15)z+(1)/(2)+2z-(6)/(5)=(5)/(12)-(1)/(2)z+()/(1) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{ 13 }{ 15 } z+ \frac{ 1 }{ 2 } +2z- \frac{ 6 }{ 5 } = \frac{ 5 }{ 12 } - \frac{ 1 }{ 2 } z+ \frac{ }{ 1 }$$

Solve for z

$z=\frac{127}{202}\approx 0.628712871$

Steps for Solving Linear Equation

Combine $\frac{13}{15}z$ and $2z$ to get $\frac{43}{15}z$.

$$\frac{43}{15}z+\frac{1}{2}-\frac{6}{5}=\frac{5}{12}-\frac{1}{2}z+\frac{1}{1}$$

Least common multiple of $2$ and $5$ is $10$. Convert $\frac{1}{2}$ and $\frac{6}{5}$ to fractions with denominator $10$.

$$\frac{43}{15}z+\frac{5}{10}-\frac{12}{10}=\frac{5}{12}-\frac{1}{2}z+\frac{1}{1}$$

Since $\frac{5}{10}$ and $\frac{12}{10}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{43}{15}z+\frac{5-12}{10}=\frac{5}{12}-\frac{1}{2}z+\frac{1}{1}$$

Subtract $12$ from $5$ to get $-7$.

$$\frac{43}{15}z-\frac{7}{10}=\frac{5}{12}-\frac{1}{2}z+\frac{1}{1}$$

Anything divided by one gives itself.

$$\frac{43}{15}z-\frac{7}{10}=\frac{5}{12}-\frac{1}{2}z+1$$

Convert $1$ to fraction $\frac{12}{12}$.

$$\frac{43}{15}z-\frac{7}{10}=\frac{5}{12}-\frac{1}{2}z+\frac{12}{12}$$

Since $\frac{5}{12}$ and $\frac{12}{12}$ have the same denominator, add them by adding their numerators.

$$\frac{43}{15}z-\frac{7}{10}=\frac{5+12}{12}-\frac{1}{2}z$$

Add $5$ and $12$ to get $17$.

$$\frac{43}{15}z-\frac{7}{10}=\frac{17}{12}-\frac{1}{2}z$$

Add $\frac{1}{2}z$ to both sides.

$$\frac{43}{15}z-\frac{7}{10}+\frac{1}{2}z=\frac{17}{12}$$

Combine $\frac{43}{15}z$ and $\frac{1}{2}z$ to get $\frac{101}{30}z$.

$$\frac{101}{30}z-\frac{7}{10}=\frac{17}{12}$$

Add $\frac{7}{10}$ to both sides.

$$\frac{101}{30}z=\frac{17}{12}+\frac{7}{10}$$

Least common multiple of $12$ and $10$ is $60$. Convert $\frac{17}{12}$ and $\frac{7}{10}$ to fractions with denominator $60$.

$$\frac{101}{30}z=\frac{85}{60}+\frac{42}{60}$$

Since $\frac{85}{60}$ and $\frac{42}{60}$ have the same denominator, add them by adding their numerators.

$$\frac{101}{30}z=\frac{85+42}{60}$$

Add $85$ and $42$ to get $127$.

$$\frac{101}{30}z=\frac{127}{60}$$

Multiply both sides by $\frac{30}{101}$, the reciprocal of $\frac{101}{30}$.

$$z=\frac{127}{60}\times \frac{30}{101}$$

Multiply $\frac{127}{60}$ times $\frac{30}{101}$ by multiplying numerator times numerator and denominator times denominator.

$$z=\frac{127\times 30}{60\times 101}$$

Do the multiplications in the fraction $\frac{127\times 30}{60\times 101}$.

$$z=\frac{3810}{6060}$$

Reduce the fraction $\frac{3810}{6060}$ to lowest terms by extracting and canceling out $30$.

$$z=\frac{127}{202}$$