Simple equation

Solution 1: By elimination method

$x+y=4$ $\color{#D61F06}(1)$

$x-y=2$ $\color{#D61F06}(2)$

Adding $\color{#D61F06}(1)$ and $\color{#D61F06}(2)$ , we get

$2x=6$

Dividing both sides by $2$ , we get $\boxed{x=3}$

Substitute $x=3$ in any of the two equations. In here I will substitute in $\color{#D61F06}(2)$ , we have

$3-y=2$

Add $y$ to both sides of the equation. We have,

$3-y+y=2+y$

$3=2+y$

Subtract $2$ from both sides of the equation. We have,

$3-2=2-2+y$

$1=y$ or $\boxed{y=1}$

$answer:(3,1)$

Solution 2: By substitution method

Solve for $x$ in terms of $y$ in $\color{#D61F06}(1)$ , then substitute in $\color{#D61F06}(2)$ . We have,

$x+y=4$ $\implies$ $x=4-y$

Substitute $x=4-y$ in $\color{#D61F06}(2)$ . We have,

$x-y=2$

$4-y-y=2$

$4-2y=2$

Subtract $4$ from both sides of the equation. We have,

$4-4-2y=2-4$

$-2y = -2$

Divide both sides by $-2$ . We get

$\boxed{y=1}$

It follows that, $x=4-y=4-1=$ $\boxed{3}$

$answer:(3,1)$

Relevant wiki: Systems of Linear Equations - Elimination

x + y = 4.....................................(1)

x - y = 2......................................(2)

From equation (2),

x = y + 2.......................................(3)

Placing the value of x to equation (1) we get, [ From equation (3)]

y + 2 + y = 4

=> 2y + 2 = 4

=> 2y = 4 - 2

=> 2y = 2

=> y = $\frac{2}{2}$

=> y = 1

Now, placing y = 1 in equation (3)

y = 1+ 2

x = 3

(x, y) = (3, 1)

Answer: (3,1)

Relevant wiki: Systems of Linear Equations - Elimination

$\begin{cases} x+y = 4 \\ x - y = 2 \end{cases}$

Adding both equations gives

$2x = 6 \implies x=3$

Subtracting the bottom equation from the top one gives

$2y = 2 \implies y=1$

Hence the solution is $\boxed{(3,1)}$ .

x + y = 4.....................................(1)

x - y = 2......................................(2)

so, 2x= 6.............................[subtracting equation 2 from equation 1]

or,x=3

now, 3+y =4......................................(1)

or, y=1

given:

x+y=4

x-y=2

First let's focus on the second equation,we need to find the value of x, to find x, transpose y to the right side of the equation we get

x=2+y

then insert the value of x to first equation making it

(2+y)+y=4

2+y+y=4

2+2y=4

transpose positive 2 to the right side of the equation making it

2y=4-2

2y=2

y=1

we get the value of y lets go back to the equation x=2+y

x=2+(1)

x=3

we find that the value of x is 3, since coordinates come in pairs (x,y) the ordered pair is (3,1)

note: to transpose terms to other side of the equation, you need to take the opposite operation in both sides of the equation.

$x + y = 4 \\ x - y = 2$ If we add the two equations together we will get $2x = 2$ , meaning that $x = 1$ .

Hence. $1 + y = 4$ , and so $y = 3$ . The solution to the simultaneous equations is $\boxed{(3, 1)}$ .

Therefore, the answer is $(3, 1)$ .

We have been given x+y=4 and only (3,1) in the options satisfy this!