Solve the following systems of equations using the elimination method: 1) 2x + y = 6, 2x − y = 2 2) 2x − y…

Question

Solve the following systems of equations using the elimination method: 1) 2x + y = 6, 2x − y = 2 2) 2x − y = 2, 2x − 5y = −6 3) x + 4y = 14, −2x + y = 10

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the systems of equations using the elimination method.

1)
Given the system of equations:

\begin{cases} 2x + y = 6 \quad &(1) \\ 2x - y = 2 \quad &(2) \end{cases}

Add equation (1) and equation (2) to eliminate $y$ .
$(2x + y) + (2x - y) = 6 + 2$ $4x = 8$
Solve for $x$ .
$x = \frac{8}{4}$ $x = 2$
Substitute the value of $x$ into equation (1) to find $y$ .
$2(2) + y = 6$ $4 + y = 6$ $y = 6 - 4$ $y = 2$
The solution is $\boxed{x=2, y=2}$ .

2)
Given the system of equations:
$\begin{cases} 2x - y = 2 \quad &(1) \\ 2x - 5y = -6 \quad &(2) \end{cases}$
Subtract equation (2) from equation (1) to eliminate $x$ .
$(2x - y) - (2x - 5y) = 2 - (-6)$ $2x - y - 2x + 5y = 2 + 6$ $4y = 8$
Solve for $y$ .
$y = \frac{8}{4}$ $y = 2$
Substitute the value of $y$ into equation (1) to find $x$ .
$2x - (2) = 2$ $2x - 2 = 2$ $2x = 2 + 2$ $2x = 4$ $x = \frac{4}{2}$ $x = 2$
The solution is $\boxed{x=2, y=2}$ .

3)
Given the system of equations (interpreting $x+4y-14$ as $x+4y=14$ ):
$\begin{cases} x + 4y = 14 \quad &(1) \\ -2x + y = 10 \quad &(2) \end{cases}$
Multiply equation (1) by 2 to make the coefficients of $x$ opposites.
$2(x + 4y) = 2(14)$ 2x + 8y = 28 \quad &(3)
Add equation (3) and equation (2) to eliminate $x$ .
$(2x + 8y) + (-2x + y) = 28 + 10$ $9y = 38$
Solve for $y$ .
$y = \frac{38}{9}$
Substitute the value of $y$ into equation (2) to find $x$ .
$-2x + \frac{38}{9} = 10$ $-2x = 10 - \frac{38}{9}$ $-2x = \frac{90}{9} - \frac{38}{9}$ $-2x = \frac{52}{9}$ $x = \frac{52}{9} \times \left(-\frac{1}{2}\right)$ $x = -\frac{26}{9}$
The solution is $\boxed{x=-\frac{26}{9}, y=\frac{38}{9}}$ .

Solve the following systems of equations using the elimination method: 1) 2x + y = 6, 2x − y = 2 2) 2x − y = 2, 2x - 5y = -6 3) x + 4y = 14, -2x + y = 10

Step-by-step solution

Add equation (1) and equation (2) to eliminate $y$ .

Solve for $x$ .

Substitute the value of $x$ into equation (1) to find $y$ .

Subtract equation (2) from equation (1) to eliminate $x$ .

Solve for $y$ .

Substitute the value of $y$ into equation (1) to find $x$ .

Multiply equation (1) by 2 to make the coefficients of $x$ opposites.

Add equation (3) and equation (2) to eliminate $x$ .

Solve for $y$ .

Substitute the value of $y$ into equation (2) to find $x$ .

Still stuck on this one?

More Mathematics Questions

Solve the following systems of equations using the elimination method: 1) 2x + y = 6, 2x − y = 2 2) 2x − y = 2, 2x - 5y = -6 3) x + 4y = 14, -2x + y = 10

Step-by-step solution

Add equation (1) and equation (2) to eliminate yyy.

Solve for xxx.

Substitute the value of xxx into equation (1) to find yyy.

Subtract equation (2) from equation (1) to eliminate xxx.

Solve for yyy.

Substitute the value of yyy into equation (1) to find xxx.

Multiply equation (1) by 2 to make the coefficients of xxx opposites.

Add equation (3) and equation (2) to eliminate xxx.

Solve for yyy.

Substitute the value of yyy into equation (2) to find xxx.

Still stuck on this one?

More Mathematics Questions

Add equation (1) and equation (2) to eliminate $y$ .

Solve for $x$ .

Substitute the value of $x$ into equation (1) to find $y$ .

Subtract equation (2) from equation (1) to eliminate $x$ .

Solve for $y$ .

Substitute the value of $y$ into equation (1) to find $x$ .

Multiply equation (1) by 2 to make the coefficients of $x$ opposites.

Add equation (3) and equation (2) to eliminate $x$ .

Solve for $y$ .

Substitute the value of $y$ into equation (2) to find $x$ .