Question

Solve for x and y
4x+y=23
3x-y=12

Answer 1

Answer: y = 3, x = 5

Explanation:

Solution by substitution method

4x+y=23

3x-y=12

Suppose,

4x+y=23→(1)

and 3x-y=12→(2)

Taking equation (1), we have

4x+y=23

⇒y= -4x+23→(3)

Putting y= -4x+23 in equation (2), we get

3x-y=12

3x-1(-4x+23)=12

⇒3x+4x-23=12

⇒7x-23=12

⇒7x=12+23

⇒7x=35

⇒x=5→(4)

Now, Putting x=5 in equation (3), we get

y=-4x+23

⇒y=-4(5)+23

⇒y=-20+23

⇒y=3

∴y=3 and x=5

Answer 2

(x,y)=(5,3)

Explanation:

Substitution Method

`\begin{bmatrix}4x+y=23\\ 3x-y=12\end{bmatrix}\\\\Isolate \:and\: solve\:for\:x :\\4x+y=23 =x=(23-y)/(4)\\\\\mathrm{Substitute\:}x=(23-y)/(4)\\\\\begin{bmatrix}3\cdot (23-y)/(4)-y=12\end{bmatrix}\\\\Simplify\\\\\begin{bmatrix}(69-7y)/(4)=12\end{bmatrix}\\\\Isolate \:and\: solve\:for\:y :\\\\(69-7y)/(4)=12\\\\y =3\\\\\mathrm{For\:}x=(23-y)/(4)\\\\\mathrm{Substitute\:}y=3\\\\x=(23-3)/(4)\\\\(23-3)/(4) =5\\\\x=5,\:y=3`

Elimination Method

`\begin{bmatrix}4x+y=23\\ 3x-y=12\end{bmatrix}\\\\\mathrm{Multiply\:}4x+y=23\mathrm{\:by\:}3\:\\\mathrm{:}\:\quad \:12x+3y=69\\\\\mathrm{Multiply\:}3x-y=12\mathrm{\:by\:}4\:\\\mathrm{:}\:\quad \:12x-4y=48\\\\\begin{bmatrix}12x+3y=69\\ 12x-4y=48\end{bmatrix}\\\\12x-4y=48\\-\\\underline{12x+3y=69}\\-7y=-21\\\\Solve :-7y=-21\\\\\mathrm{Divide\:both\:sides\:by\:}-7\\(-7y)/(-7)=(-21)/(-7)\\\\Simplify\\y=3\\\\\mathrm{For\:}12x+3y=69,\\\mathrm{\:plug\:in\:}y=3\\`

`Solve :12x+3\cdot \:3=69\\\\12x+9=69\\\\12x+9-9=69-9\\\\Simplify\\\\12x=60\\\\\mathrm{Divide\:both\:sides\:by\:}12\\\\(12x)/(12)=(60)/(12)\\\\Simplify\\\\x=5\\\\\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}\\\\x=5,\:y=3`