Solve the Linear System: -x + y = 14 and 5x + 2y = 7

Q: Why multiply the first equation by 2 instead of the second equation?

We chose to multiply the first equation by 2 because it's simpler! This makes the y-coefficients both equal to 2, so they cancel perfectly when we subtract. You could also multiply the second equation by different numbers, but this way is more efficient.

Q: Can I eliminate x instead of y?

Absolutely! You could multiply the first equation by 5 to get $ -5x + 5y = 70 $, then add it to the second equation. Both methods work - choose whichever looks easier!

Q: What if I get different coefficients that don't match easily?

Sometimes you need to multiply both equations by different numbers. For example, if you had 3x and 5x, multiply the first by 5 and the second by 3 to get 15x in both equations.

Q: How do I check if my solution is correct?

Substitute your values into both original equations:First: $ -(-3) + 11 = 3 + 11 = 14 $ ✓Second: $ 5(-3) + 2(11) = -15 + 22 = 7 $ ✓

Q: What if I get a negative answer? Is that wrong?

Negative answers are completely normal! In this problem, $ x = -3 $ is correct. Always trust your algebra - if you followed the steps correctly, negative solutions are just as valid as positive ones.

Linear Systems with Elimination Method

$-x+y=14$

$5x+2y=7$

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Understand the problem

$-x+y=14$

$5x+2y=7$

Step-by-step solution

To solve this system of equations, we'll employ the elimination method.

The given system is:

$-x + y = 14$
$5x + 2y = 7$

To eliminate $y$ , we can multiply the first equation by $2$ to match the coefficient of $y$ in the second equation:

$2(-x + y) = 2 \times 14$

Resulting in:

$-2x + 2y = 28$

Now, we have:

$-2x + 2y = 28$
$5x + 2y = 7$

Subtract the second equation from the first:

$\begin{aligned} (-2x + 2y) - (5x + 2y) &= 28 - 7 \\ -2x + 2y - 5x - 2y &= 21 \\ -7x &= 21 \end{aligned}$

Solving for $x$ :

$x = \frac{21}{-7} = -3$

Next, substitute $x = -3$ back into the first equation:

$-(-3) + y = 14$

$3 + y = 14$

Solving for $y$ :

$y = 14 - 3 = 11$

Therefore, the solution to the system of equations is $x = -3$ and $y = 11$ .

The correct answer choice is:

$x=-3,y=11$

Final Answer

$x=-3,y=11$

Key Points to Remember

Essential concepts to master this topic

Rule: Match coefficients by multiplying equations before eliminating variables
Technique: Multiply first equation by 2: $-2x + 2y = 28$
Check: Substitute $x = -3, y = 11$ into both original equations ✓

Common Mistakes

Avoid these frequent errors

Adding equations without matching coefficients first
Don't directly add $-x + y = 14$ and $5x + 2y = 7$ = variables won't cancel out! This leaves you with messy coefficients that don't eliminate cleanly. Always multiply one or both equations to match coefficients before adding or subtracting.

Practice Quiz

Test your knowledge with interactive questions

$\begin{cases} x+y=8 \\ x-y=6 \end{cases}$

FAQ

Everything you need to know about this question

Why multiply the first equation by 2 instead of the second equation?

We chose to multiply the first equation by 2 because it's simpler! This makes the y-coefficients both equal to 2, so they cancel perfectly when we subtract. You could also multiply the second equation by different numbers, but this way is more efficient.

Can I eliminate x instead of y?

Absolutely! You could multiply the first equation by 5 to get $-5x + 5y = 70$ , then add it to the second equation. Both methods work - choose whichever looks easier!

What if I get different coefficients that don't match easily?

Sometimes you need to multiply both equations by different numbers. For example, if you had 3x and 5x, multiply the first by 5 and the second by 3 to get 15x in both equations.

How do I check if my solution is correct?

Substitute your values into both original equations:

First: $-(-3) + 11 = 3 + 11 = 14$ ✓
Second: $5(-3) + 2(11) = -15 + 22 = 7$ ✓

What if I get a negative answer? Is that wrong?

Negative answers are completely normal! In this problem, $x = -3$ is correct. Always trust your algebra - if you followed the steps correctly, negative solutions are just as valid as positive ones.

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