Solve 1/4left(6x-7/40-3right)=4x-7/40+1

$$\frac{ 1 }{ 4 } \left( 6x- \frac{ 7 }{ 40 } -3 \right) = 4x \frac{ -7 }{ 40 } +1$$

Solve for x

$$x=\frac{287}{352}\approx 0.815340909$$

Steps for Solving Linear Equation

Convert 3 to fraction \frac{120}{40}.

$$\frac{1}{4}\left(6x-\frac{7}{40}-\frac{120}{40}\right)=4x\times \frac{-7}{40}+1$$

Since -\frac{7}{40} and \frac{120}{40} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{4}\left(6x+\frac{-7-120}{40}\right)=4x\times \frac{-7}{40}+1$$

Subtract 120 from -7 to get -127.

$$\frac{1}{4}\left(6x-\frac{127}{40}\right)=4x\times \frac{-7}{40}+1$$

Use the distributive property to multiply \frac{1}{4} by 6x-\frac{127}{40}.

$$\frac{1}{4}\times 6x+\frac{1}{4}\left(-\frac{127}{40}\right)=4x\times \frac{-7}{40}+1$$

Multiply \frac{1}{4} and 6 to get \frac{6}{4}.

$$\frac{6}{4}x+\frac{1}{4}\left(-\frac{127}{40}\right)=4x\times \frac{-7}{40}+1$$

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

$$\frac{3}{2}x+\frac{1}{4}\left(-\frac{127}{40}\right)=4x\times \frac{-7}{40}+1$$

Multiply \frac{1}{4} times -\frac{127}{40} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3}{2}x+\frac{1\left(-127\right)}{4\times 40}=4x\times \frac{-7}{40}+1$$

Do the multiplications in the fraction \frac{1\left(-127\right)}{4\times 40}.

$$\frac{3}{2}x+\frac{-127}{160}=4x\times \frac{-7}{40}+1$$

Fraction \frac{-127}{160} can be rewritten as -\frac{127}{160} by extracting the negative sign.

$$\frac{3}{2}x-\frac{127}{160}=4x\times \frac{-7}{40}+1$$

Fraction \frac{-7}{40} can be rewritten as -\frac{7}{40} by extracting the negative sign.

$$\frac{3}{2}x-\frac{127}{160}=4x\left(-\frac{7}{40}\right)+1$$

Express 4\left(-\frac{7}{40}\right) as a single fraction.

$$\frac{3}{2}x-\frac{127}{160}=\frac{4\left(-7\right)}{40}x+1$$

Multiply 4 and -7 to get -28.

$$\frac{3}{2}x-\frac{127}{160}=\frac{-28}{40}x+1$$

Reduce the fraction \frac{-28}{40} to lowest terms by extracting and canceling out 4.

$$\frac{3}{2}x-\frac{127}{160}=-\frac{7}{10}x+1$$

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

$$\frac{3}{2}x-\frac{127}{160}+\frac{7}{10}x=1$$

Combine \frac{3}{2}x and \frac{7}{10}x to get \frac{11}{5}x.

$$\frac{11}{5}x-\frac{127}{160}=1$$

Add \frac{127}{160} to both sides.

$$\frac{11}{5}x=1+\frac{127}{160}$$

Convert 1 to fraction \frac{160}{160}.

$$\frac{11}{5}x=\frac{160}{160}+\frac{127}{160}$$

Since \frac{160}{160} and \frac{127}{160} have the same denominator, add them by adding their numerators.

$$\frac{11}{5}x=\frac{160+127}{160}$$

Add 160 and 127 to get 287.

$$\frac{11}{5}x=\frac{287}{160}$$

Multiply both sides by \frac{5}{11}, the reciprocal of \frac{11}{5}.

$$x=\frac{287}{160}\times \frac{5}{11}$$

Multiply \frac{287}{160} times \frac{5}{11} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{287\times 5}{160\times 11}$$

Do the multiplications in the fraction \frac{287\times 5}{160\times 11}.

$$x=\frac{1435}{1760}$$

Reduce the fraction \frac{1435}{1760} to lowest terms by extracting and canceling out 5.

$$x=\frac{287}{352}$$