Question

Y=x+14 x+2y=36 simultaneous equation. how to solve+ working pls

Answer 1

`\tt \bigg((8)/(3), \: (50)/(3)\bigg)`

Explanation:

Given equations are,

`{\implies \sf y = x+14 \qquad - \big\lgroup eq^n \:1 \big\rgroup}`

`{\implies\sf x+2y=36 \qquad - \big\lgroup eq^n \:2 \big\rgroup}`

We have the value of y as x + 14. Let us substitute the value of y in eqⁿ 1. Then we get,

`{\implies \sf x + 2(x+14) = 36}`

• Solve the bracket.

`{\implies \sf x + 2x+28 = 36}`

• Combining like terms (x and 2x)

`{\implies \sf 3x+28 = 36}`

• Subtracting 28 from both sides

`{\implies \sf 3x+28-28 = 36-28}`

`{\implies \sf 3x = 8}`

• Divide both sides by 3

`{\implies \boxed{\sf x= (8)/(3)}}`

Now, put the value of x in eqⁿ 1.

`{\implies \sf y = x+14}`

`{\implies \sf y = (8)/(3)+14}`

`{\implies \sf y = (8)/(3)+(14)/(1)}`

`{\implies \sf y = (8)/(3)+(14* 3)/(1* 3)}`

`{\implies \sf y = (8)/(3)+(42)/(3)}`

`{\implies \sf y = (8+42)/(3)}`

`{\implies \boxed{\sf y = (50)/(3)}}`

Therefore,

`\sf \bigg((8)/(3) , (50)/(3) \bigg) \:is \:the \:solution \:for \:the\:given \:equation.`

Answer 2

Explanation:

we have
`\left \{ {{y=x+14} \atop {x+2y=36}} \right.`

⇒
`\left \{ {{x=y-14} \atop {x=-2y+36}} \right.`

⇒y-14 = -2y+36

⇒3y = 50

⇒y =
`(50)/(3)`

⇒x = y-14 =
`(8)/(3)`