Question

Question #17602

Answer 1

Use the elimination or substitution methods

Explanation:

Here I would use the elimination method and get rid of the y variable. The first equation can be multiplied by 3 to make the equation `-9x-6y=-15`

Then add the two equations together to get rid of the y variable and solve for the `x`
`-9x-6y=-15`
`ul(+ 7x+6y=+1 larr" Add")`
`-2x+color(white)(".")0=-14`

`=>color(red)(x=7)`

Plug in for `y`

`7( color(red)(7) )+6y=1`

`49+6y=1`

`6y=-48`

`y=-8`

Double check by plugging in both solutions
The answer is (7, -8)

Answer 2

`x=7`
`y=-8`

Explanation:

`{(-3x-2y=-5),(7x+6y=1):}`

Let's procede finding the x of the second equation:
`{ (-3x-2y=-5),(7x=-6y+1):}`

`{(-3x-2y=-5),(x=-\frac{6}{7}y+\frac{1}{7}):}`

Now we can substitute the x that we found in the first equation:
`{(-3(-\frac{6}{7}y+\frac{1}{7})-2y=-5),(x=-\frac{6}{7}y+\frac{1}{7}):}`

`{(\frac{18}{7}y-\frac{3}{7}-2y=-5),(x=-\frac{6}{7}y+\frac{1}{7}):}`

`{(\frac{18}{7}y-2y=-5+\frac{3}{7}),(x=-\frac{6}{7}y+\frac{1}{7}):}`

#{(\frac{18-14}{\cancel{7}}y=\frac{3-35}{\cancel{7}}), (x=-\frac{6}{7}y+\frac{1}{7}):}#

`{(4y=-32),(x=-\frac{6}{7}y+\frac{1}{7}):}`

`{(y=-8),(x=-\frac{6}{7}y+\frac{1}{7}):}`

Now let's substitute `y=-8` in the second equation:
`{(y=-8),(x=-\frac{6}{7}(-8)+\frac{1}{7}):}`

`{(y=-8),(x=\frac{48}{7}+\frac{1}{7}):}`

`{(y=-8),(x=\frac{49}{7}=7):}`

Final result:
`x=7`
`y=-8`

Answer 3

Have a look at https://socratic.org/help/symbols Note the hash symbols at the beginning and end of the entered text.

`x=7` and `y=-8`

Explanation:

Given:
`color(green)(-3x-2y=-5" "....................Equation(1))`
`color(green)(+7x+6y=+1" "....................Equation(2))`

`[7xxEqn(1)]+[3xxEqn(2)]`
`color(white)("dddddddddd") darr`

`color(green)(-21x-14y=-35" "....Equation(1_a))`
`color(green)(ul(+21x+18y=+color(white)(5)3)........Equation(2_a))`
`color(green)(color(white)("dd")0color(white)("d.")+color(white)("d")4y=-32)`

`color(green)( y=-8 )`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I choose `Eqn(2)`
Substitute for `color(red)(y)` in `Eqn(2)` giving:

`color(green)(+7x+6color(red)(y)=+1color(white)("ddd")->color(white)("ddd")7x+[6color(red)(xx(-8))]=1)`

`color(white)("ddddddddddddddd.d")color(green)(->color(white)("ddd")7x-color(white)("ddd")48color(white)("dddd")=1)`

`color(white)("ddddddddddddddd.d")color(green)(->color(white)("ddddddddddddd.d")7x=49)`

`color(green)(color(white)("ddddddddddddddddd")->color(white)("ddddddddddddddd") x=49/7=7 )`