Solve the system 2(x - y) + x = 95 and (x - y) + 3(x + y) = 150 for x and y.

Step-by-step solution

1. 1

   Expand the brackets before choosing a method. Neither equation is in standard form, so elimination or substitution cannot be applied yet. Distributing in the first equation:

   $2(x-y)+x=2x-2y+x=3x-2y,$

   so equation one is $3x-2y=95$. Trying to eliminate before expanding is the usual reason these systems go wrong.

2. 2

   Expand and simplify the second equation. Distributing the $3$:

   $(x-y)+3(x+y)=x-y+3x+3y=4x+2y,$

   so the second equation is $4x+2y=150$. Every coefficient is even, so divide through by $2$:

   $2x+y=75.$

   Reducing now keeps all later arithmetic small — a genuinely useful habit, not just cosmetics.

3. 3

   Choose substitution because one coefficient is 1. The system is

   $3x-2y=95,\qquad 2x+y=75.$

   The $y$ in the second equation has coefficient $1$, so solving for it introduces no fractions:

   $y=75-2x.$

4. 4

   Substitute and solve for $x$. Putting $y=75-2x$ into $3x-2y=95$:

   $3x-2(75-2x)=95\;\Longrightarrow\;3x-150+4x=95\;\Longrightarrow\;7x=245\;\Longrightarrow\;x=35.$

   The sign detail is $-2\times(-2x)=+4x$: two negatives make the $x$ terms add rather than cancel.

5. 5

   Back-substitute and check in the original equations. From $2x+y=75$ with $x=35$: $70+y=75$, so $y=5$. Checking against the untouched originals rather than the simplified ones:

   $2(35-5)+35=60+35=95 ✓,\qquad (35-5)+3(35+5)=30+120=150 ✓.$

   Verifying in the originals is what catches an expansion mistake made back in step 1.