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Solve the system using elimination.

[tex]
\begin{array}{l}
2x + 3y = 8 \\
5x + 6y = 17
\end{array}
[/tex]


Sagot :

Sure, let's solve the given system of equations using the elimination method. The system of equations is:

[tex]\[ \begin{cases} 2x + 3y = 8 \\ 5x + 6y = 17 \end{cases} \][/tex]

### 1. Eliminate one of the variables

We can start by eliminating [tex]\( y \)[/tex]. To do this, we want to make the coefficients of [tex]\( y \)[/tex] in both equations the same. We can achieve this by multiplying the entire first equation by 2, so that the coefficient of [tex]\( y \)[/tex] in the first equation becomes 6:

[tex]\[ 2(2x + 3y) = 2(8) \][/tex]

This simplifies to:

[tex]\[ 4x + 6y = 16 \][/tex]

Now our system of equations is:

[tex]\[ \begin{cases} 4x + 6y = 16 \\ 5x + 6y = 17 \end{cases} \][/tex]

### 2. Subtract the equations to eliminate [tex]\( y \)[/tex]

Now we subtract the first modified equation from the second original equation to eliminate [tex]\( y \)[/tex]:

[tex]\[ (5x + 6y) - (4x + 6y) = 17 - 16 \][/tex]

Simplifying this, we get:

[tex]\[ 5x + 6y - 4x - 6y = 1 \][/tex]

[tex]\[ x = 1 \][/tex]

So, we have found [tex]\( x = 1 \)[/tex].

### 3. Find the value of [tex]\( y \)[/tex]

Next, we substitute [tex]\( x = 1 \)[/tex] back into one of the original equations to find [tex]\( y \)[/tex]. Let's use the first equation:

[tex]\[ 2x + 3y = 8 \][/tex]

Substitute [tex]\( x = 1 \)[/tex]:

[tex]\[ 2(1) + 3y = 8 \][/tex]

This simplifies to:

[tex]\[ 2 + 3y = 8 \][/tex]

Subtract 2 from both sides:

[tex]\[ 3y = 6 \][/tex]

Divide both sides by 3:

[tex]\[ y = 2 \][/tex]

### Conclusion

The solution to the system of equations is [tex]\( x = 1 \)[/tex] and [tex]\( y = 2 \)[/tex]. So, the solution is [tex]\((1, 2)\)[/tex].