Q. If (x – 12y = -210) and (x – 6y = 90), then (x =)
Answer
\[
\begin{aligned}
x-12y&=-210\\
x-6y&=90
\end{aligned}
\]
Subtract the second equation from the first to eliminate \(x\):
\[
(x-12y)-(x-6y)=-210-90
\]
\[
-6y=-300
\]
\[
y=50
\]
Substitute \(y=50\) into the second equation:
\[
x-6(50)=90
\]
\[
x-300=90
\]
\[
x=390
\]
Final solution:
\[
\boxed{(x,y)=(390,50)}
\]
Detailed Explanation
Problem: Solve the system
\( x – 12y = -210 \) and \( x – 6y = 90 \).
Step 1 — Eliminate \(x\)
Subtract the second equation from the first to eliminate \(x\):
\[
(x-12y)-(x-6y)=-210-90
\]
Simplify the left-hand side (the \(x\) terms cancel):
\[
x-12y-x+6y=-6y
\]
So we get
\[
-6y=-300
\]
Step 2 — Solve for \(y\)
Divide both sides by \(-6\):
\[
y=\frac{-300}{-6}=50
\]
Step 3 — Substitute to find \(x\)
Substitute \(y=50\) into \(x-6y=90\):
\[
x-6(50)=90
\]
\[
x-300=90
\]
\[
x=90+300=390
\]
Answer
\[
(x,y)=(390,50)
\]
See full solution
FAQs
How do I find x from the system (x - 12y = -210) and (x - 6y = 90)?
Subtract the second equation from the first to eliminate (x): (-6y=-300) so (y=50). Substitute into (x-6y=90): (x-300=90), hence (x=390).
Can I use elimination to solve this system?
Yes. Subtract the equations to eliminate (x) (since both have coefficient 1). That yields (-6y=-300), (y=50); then substitute to get (x=390).
How would substitution work here?
From (x-6y=90) get (x=90+6y). Plug into (x-12y=-210): ((90+6y)-12y=-210) → (-6y=-300) → (y=50); then (x=90+6(50)=390).
Is the solution unique, infinite, or none?
Unique. Two non-identical linear equations in two variables intersect at a single point, here ((x,y)=(390,50)).
What is the geometric interpretation of the solution?
Each equation is a line in the plane; their intersection is the single point ((390,50)), meaning both lines meet exactly there.
How can I quickly check my answer?
Substitute (x=390) and (y=50) into both: (390-12(50)=390-600=-210) and (390-6(50)=390-300=90). Both true, so correct.
What common mistakes should I avoid?
Watch sign changes when subtracting equations and avoid arithmetic slips. Also remember to substitute the correct (y) value back to find (x).
If the constants had been different so subtraction gave 0, what then?
If subtraction yields (0=0), the lines are identical (infinitely many solutions). If it yields a contradiction like (0=c) with (cneq0), there is no solution (parallel lines).
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