Question

Solving systems by elimination 2x-10y=30 10x-y=19

Answer 1

To solve the system of equations using the elimination method, we start with the following equations:

1. \( 2x - 10y = 30 \) (Equation 1)
2. \( 10x - y = 19 \) (Equation 2)

First, let's manipulate these equations to align the coefficients for elimination. We will eliminate \( y \).

We can multiply Equation 2 by 10 to make the coefficients of \( y \) in both equations the same:

\[
10(10x - y) = 10(19) \implies 100x - 10y = 190 \quad (Equation 3)
\]

Now we have:

1. \( 2x - 10y = 30 \) (Equation 1)
2. \( 100x - 10y = 190 \) (Equation 3)

Next, we will subtract Equation 1 from Equation 3:

\[
(100x - 10y) - (2x - 10y) = 190 - 30
\]

This simplifies to:

\[
100x - 2x = 190 - 30
\]
\[
98x = 160
\]
\[
x = \frac{160}{98} = \frac{80}{49}
\]

Now that we have \( x \), we will substitute this value back into one of the original equations to find \( y \). Let's use Equation 2:

\[
10x - y = 19
\]

Substituting \( x \):

\[
10\left(\frac{80}{49}\right) - y = 19
\]
\[
\frac{800}{49} - y = 19
\]

Now, we can isolate \( y \):

\[
-y = 19 - \frac{800}{49}
\]

To combine these, convert 19 to a fraction with a denominator of 49:

\[
19 = \frac{931}{49}
\]

So:

\[
-y = \frac{931}{49} - \frac{800}{49}
\]
\[
-y = \frac{131}{49}
\]
\[
y = -\frac{131}{49}
\]

Thus, the solution to the system of equations is:

\[
x = \frac{80}{49}, \quad y = -\frac{131}{49}
\]

In conclusion, the solution is:

\[
\left(x, y\right) = \left(\frac{80}{49}, -\frac{131}{49}\right)
\]