Question

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

Answer 1

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:

`3y=1-x`

`x = 1-3y`

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

`6(1-3y)=29-8y`

`8y-18y = 29 - 6`

`-10y = 23`

`y = -2.3`

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

`x = 1- 3y = 1-3*(-2.3) = 7.9`

check:

eq1:

`6x = 29 - 8y`

`6*(7.9) = 29-8*(-2.3)`

`47.4 = 47.4 \implies true`

eq2:

`3y = 1-x`

`3*(-2.3) = 1-7.9`

`-6.9 = -6.9 \implies true`

Answer 2

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

`color(green)(y=-23/10`

Explanation:

Given:

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

Write the equations separately:

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

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

Consider

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

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

`6x + color(blue)(8y)=29-8y+color(blue)(8y)`

`6x + color(blue)(8y)=29cancel(-8y)color(blue)cancel(+8y)`

`color(blue)(6x + 8y = 29)` Equation (3)

Consider

`3y = 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)`

`color(blue)(x + 3y = 1)` Equation (4)

Consider:

`color(blue)(6x + 8y = 29)` Equation (3)

`color(blue)(x + 3y = 1)` Equation (4)

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

This resultant equation is Equation (5)

`color(blue)(6x + 8y = 29)` Equation (3)

`color(blue)(6x + 18y = 6)` Equation (5)

Subtract Equation (5) from Equation (3)

`cancel(6x) + 8y = 29` Equation (3)

`-cancel(6x) - 18y = -6` Equation (5)

You get

`-10y=23`

Divide both sides by 10 to get

`(-cancel(10)y)/cancel(10)=23/10`

`=-y = 23/10`

Multiply both sides by `(-1)` to get

`=(-y)(-1) = (23/10)(-1)`

`color(green)(y=-23/10`

Substitute this value of `y` in `color(blue)(x + 3y = 1)` Equation (4) to get

`x+3(-23/10) = 1`

`rArr x-69/10 = 1`

Add `69/10` to both the sides to get

`rArr x-69/10 + 69/10 = 1+69/10`

`rArr x-cancel(69/10) + cancel(69/10) = 1+69/10`

`rArr x = 1+69/10`

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

`rArr x = (10+69)/10`

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

Hence, values of `x and y` are:

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

`color(green)(y=-23/10`

The system of linear equations is solved.