Answer :
Your question is -
x + y = 10
x - y = 5
Now, let us solve this system by the substitution method.
According to equation 2 ,
x - y = 5
thus, x = 5 + y
Substituting the value of x in equation 1,
(5 + y) + y = 10
5 + 2y = 10
2y = 10 - 5 = 5
thus, y = 5/2
Substituting the value of y in value of x,
x = 5 + y
= 5 + 5/2
= 15/2
Thus, x = 15/2 ( or 7.5) y = 5/2 ( or 2.5)
x + y = 10
x - y = 5
Now, let us solve this system by the substitution method.
According to equation 2 ,
x - y = 5
thus, x = 5 + y
Substituting the value of x in equation 1,
(5 + y) + y = 10
5 + 2y = 10
2y = 10 - 5 = 5
thus, y = 5/2
Substituting the value of y in value of x,
x = 5 + y
= 5 + 5/2
= 15/2
Thus, x = 15/2 ( or 7.5) y = 5/2 ( or 2.5)
[tex]\begin{cases} x + y = 10 \\ x - y = 5 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ x - y = 5 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ x - (10-x) = 5 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ x - 10+x = 5 \end{cases}\\ \\[/tex]
[tex]\begin{cases} y = 10 -x\\ 2x - 10 = 5 \end{cases}\\ \\\begin{cases} y = 10 -x\\ 2x = 5 +10 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ 2x =15 \ \ /:2 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ x =\frac{15}{2} \end{cases}\\ \\[/tex]
[tex]\begin{cases} y = 10 - 7\frac{1}{2}\\ x =7\frac{1}{2} \end{cases}\\ \\ \begin{cases} y = 2\frac{1}{2}\\ x = 8\frac{1}{2} \end{cases}\\ \\[/tex]
[tex]\begin{cases} y = 10 -x\\ 2x - 10 = 5 \end{cases}\\ \\\begin{cases} y = 10 -x\\ 2x = 5 +10 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ 2x =15 \ \ /:2 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ x =\frac{15}{2} \end{cases}\\ \\[/tex]
[tex]\begin{cases} y = 10 - 7\frac{1}{2}\\ x =7\frac{1}{2} \end{cases}\\ \\ \begin{cases} y = 2\frac{1}{2}\\ x = 8\frac{1}{2} \end{cases}\\ \\[/tex]