One equation of a pair of dependent linear equation 3x 8y 5 the second equation can be

2x-5y-43x=-8y

Consider the first equation. Subtract 43x from both sides.

-41x-5y=-8y

Combine 2x and -43x to get -41x.

-41x-5y+8y=0

Add 8y to both sides.

-41x+3y=0

Combine -5y and 8y to get 3y.

-41x+3y=0,43x-8y=5

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

-41x+3y=0

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

-41x=-3y

Subtract 3y from both sides of the equation.

x=-\frac{1}{41}\left(-3\right)y

Divide both sides by -41.

x=\frac{3}{41}y

Multiply -\frac{1}{41} times -3y.

43\times \left(\frac{3}{41}\right)y-8y=5

Substitute \frac{3y}{41} for x in the other equation, 43x-8y=5.

\frac{129}{41}y-8y=5

Multiply 43 times \frac{3y}{41}.

-\frac{199}{41}y=5

Add \frac{129y}{41} to -8y.

y=-\frac{205}{199}

Divide both sides of the equation by -\frac{199}{41}, which is the same as multiplying both sides by the reciprocal of the fraction.

x=\frac{3}{41}\left(-\frac{205}{199}\right)

Substitute -\frac{205}{199} for y in x=\frac{3}{41}y. Because the resulting equation contains only one variable, you can solve for x directly.

x=-\frac{15}{199}

Multiply \frac{3}{41} times -\frac{205}{199} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

x=-\frac{15}{199},y=-\frac{205}{199}

The system is now solved.

2x-5y-43x=-8y

Consider the first equation. Subtract 43x from both sides.

-41x-5y=-8y

Combine 2x and -43x to get -41x.

-41x-5y+8y=0

Add 8y to both sides.

-41x+3y=0

Combine -5y and 8y to get 3y.

-41x+3y=0,43x-8y=5

Put the equations in standard form and then use matrices to solve the system of equations.

\left(\begin{matrix}-41&3\\43&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\5\end{matrix}\right)

Write the equations in matrix form.

inverse(\left(\begin{matrix}-41&3\\43&-8\end{matrix}\right))\left(\begin{matrix}-41&3\\43&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-41&3\\43&-8\end{matrix}\right))\left(\begin{matrix}0\\5\end{matrix}\right)

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-41&3\\43&-8\end{matrix}\right).

\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-41&3\\43&-8\end{matrix}\right))\left(\begin{matrix}0\\5\end{matrix}\right)

The product of a matrix and its inverse is the identity matrix.

\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-41&3\\43&-8\end{matrix}\right))\left(\begin{matrix}0\\5\end{matrix}\right)

Multiply the matrices on the left hand side of the equal sign.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{-8}{-41\left(-8\right)-3\times 43}&-\frac{3}{-41\left(-8\right)-3\times 43}\\-\frac{43}{-41\left(-8\right)-3\times 43}&\frac{-41}{-41\left(-8\right)-3\times 43}\end{matrix}\right)\left(\begin{matrix}0\\5\end{matrix}\right)

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{199}&-\frac{3}{199}\\-\frac{43}{199}&-\frac{41}{199}\end{matrix}\right)\left(\begin{matrix}0\\5\end{matrix}\right)

Do the arithmetic.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{199}\times 5\\-\frac{41}{199}\times 5\end{matrix}\right)

Multiply the matrices.

\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{15}{199}\\-\frac{205}{199}\end{matrix}\right)

Do the arithmetic.

x=-\frac{15}{199},y=-\frac{205}{199}

Extract the matrix elements x and y.

2x-5y-43x=-8y

Consider the first equation. Subtract 43x from both sides.

-41x-5y=-8y

Combine 2x and -43x to get -41x.

-41x-5y+8y=0

Add 8y to both sides.

-41x+3y=0

Combine -5y and 8y to get 3y.

-41x+3y=0,43x-8y=5

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

43\left(-41\right)x+43\times 3y=0,-41\times 43x-41\left(-8\right)y=-41\times 5

To make -41x and 43x equal, multiply all terms on each side of the first equation by 43 and all terms on each side of the second by -41.

-1763x+129y=0,-1763x+328y=-205

Simplify.

-1763x+1763x+129y-328y=205

Subtract -1763x+328y=-205 from -1763x+129y=0 by subtracting like terms on each side of the equal sign.

129y-328y=205

Add -1763x to 1763x. Terms -1763x and 1763x cancel out, leaving an equation with only one variable that can be solved.

-199y=205

Add 129y to -328y.

y=-\frac{205}{199}

Divide both sides by -199.

43x-8\left(-\frac{205}{199}\right)=5

Substitute -\frac{205}{199} for y in 43x-8y=5. Because the resulting equation contains only one variable, you can solve for x directly.

43x+\frac{1640}{199}=5

Multiply -8 times -\frac{205}{199}.

43x=-\frac{645}{199}

Subtract \frac{1640}{199} from both sides of the equation.

x=-\frac{15}{199}

Divide both sides by 43.

x=-\frac{15}{199},y=-\frac{205}{199}

The system is now solved.

What is the equation of dependent linear equation?

The second equation can be, a. 10x + 14y + 4 = 0, b.

Is 5x 7y =

Answer. yes they are pair of dependent linear equation.

Can a pair of linear equations have 2 solutions?

No. A system of linear equations in two variables may have zero, one, or infinitely many solutions.

When the pair of linear equations is dependent?

A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables.