# How do solve the following linear system?:  6x=29-8y , 3y = 1 -x ?

Mar 9, 2018

The solution is a point at $x = 7.9$ and $y = - 2.3$

#### Explanation:

Substitution implies that we solve for either $x$ or $y$ in one equation and substitute that into the other. I see a lonely $x$ in the second equation that should be easy to solve for, so I'll start there:

$3 y = 1 - x$

$x = 1 - 3 y$

Then we substitute $1 - 3 y$ for $x$ in the first equation:

$6 \left(1 - 3 y\right) = 29 - 8 y$

$8 y - 18 y = 29 - 6$

$- 10 y = 23$

$y = - 2.3$

Then we can use our original solution for $x$ to get

$x = 1 - 3 y = 1 - 3 \cdot \left(- 2.3\right) = 7.9$

check:

eq1:

$6 x = 29 - 8 y$

$6 \cdot \left(7.9\right) = 29 - 8 \cdot \left(- 2.3\right)$

$47.4 = 47.4 \setminus \implies t r u e$

eq2:

$3 y = 1 - x$

$3 \cdot \left(- 2.3\right) = 1 - 7.9$

$- 6.9 = - 6.9 \setminus \implies t r u e$

Mar 10, 2018

color(green)( x = (79)/10

color(green)(y=-23/10

#### Explanation:

Given:

color(red)(6x=29-8y , 3y = 1 -x

Write the equations separately:

$6 x = 29 - 8 y$ Equation (1)

$3 y = 1 - x$ Equation (2)

Consider

$6 x = 29 - 8 y$ Equation (1)

Add color(blue)(8y to both sides of the equation.

$6 x + \textcolor{b l u e}{8 y} = 29 - 8 y + \textcolor{b l u e}{8 y}$

$6 x + \textcolor{b l u e}{8 y} = 29 \cancel{- 8 y} \textcolor{b l u e}{\cancel{+ 8 y}}$

$\textcolor{b l u e}{6 x + 8 y = 29}$ Equation (3)

Consider

$3 y = 1 - x$ Equation (2)

Add color(blue)(x to both sides of the equation.

3y + color(blue)(x) = 1 -x+color(blue)(x

3y + color(blue)(x) = 1 -cancel(x)+color(blue)(cancel(x)

$\textcolor{b l u e}{x + 3 y = 1}$ Equation (4)

Consider:

$\textcolor{b l u e}{6 x + 8 y = 29}$ Equation (3)

$\textcolor{b l u e}{x + 3 y = 1}$ Equation (4)

Multiply both sides of the Equation (4) by 6.

This resultant equation is Equation (5)

$\textcolor{b l u e}{6 x + 8 y = 29}$ Equation (3)

$\textcolor{b l u e}{6 x + 18 y = 6}$ Equation (5)

Subtract Equation (5) from Equation (3)

$\cancel{6 x} + 8 y = 29$ Equation (3)

$- \cancel{6 x} - 18 y = - 6$ Equation (5)

You get

$- 10 y = 23$

Divide both sides by 10 to get

$\frac{- \cancel{10} y}{\cancel{10}} = \frac{23}{10}$

$= - y = \frac{23}{10}$

Multiply both sides by $\left(- 1\right)$ to get

$= \left(- y\right) \left(- 1\right) = \left(\frac{23}{10}\right) \left(- 1\right)$

color(green)(y=-23/10

Substitute this value of $y$ in $\textcolor{b l u e}{x + 3 y = 1}$ Equation (4) to get

$x + 3 \left(- \frac{23}{10}\right) = 1$

$\Rightarrow x - \frac{69}{10} = 1$

Add $\frac{69}{10}$ to both the sides to get

$\Rightarrow x - \frac{69}{10} + \frac{69}{10} = 1 + \frac{69}{10}$

$\Rightarrow x - \cancel{\frac{69}{10}} + \cancel{\frac{69}{10}} = 1 + \frac{69}{10}$

$\Rightarrow x = 1 + \frac{69}{10}$

Take the common denominator of the numeric expression at the Right-Hand Side(RHS) to simplify:

$\Rightarrow x = \frac{10 + 69}{10}$

color(green)( x = (79)/10

Hence, values of $x \mathmr{and} y$ are:

color(green)( x = (79)/10

color(green)(y=-23/10

The system of linear equations is solved.