Question

for each ordered pair, determine whether it is a solution to $7x + 3y = 2$.

$(x,y)$	is it a solution?
$(0, -8)$	$circ$	$circ$
$(2, -4)$	$circ$	$circ$
$(-1, 3)$	$circ$	$circ$
$(5, 2)$	$circ$	$circ$

Explanation:

To determine if an ordered pair \((x, y)\) is a solution to the equation \(7x + 3y = 2\), we substitute the values of \(x\) and \(y\) into the equation and check if the left - hand side (LHS) equals the right - hand side (RHS, which is 2).

Step 1: Check for \((0,-8)\)

Substitute \(x = 0\) and \(y=-8\) into \(7x + 3y\).
The formula for substitution is \(7x+3y=7\times x + 3\times y\).
So, \(7(0)+3(-8)=0 - 24=-24\).
Since \(-24
eq2\), \((0, - 8)\) is not a solution.

Step 2: Check for \((2,-4)\)

Substitute \(x = 2\) and \(y = - 4\) into \(7x+3y\).
We have \(7(2)+3(-4)=14-12 = 2\).
Since \(2 = 2\), \((2,-4)\) is a solution.

Step 3: Check for \((-1,3)\)

Substitute \(x=-1\) and \(y = 3\) into \(7x + 3y\).
We get \(7(-1)+3(3)=-7 + 9=2\).
Since \(2=2\), \((-1,3)\) is a solution.

Step 4: Check for \((5,2)\)

Substitute \(x = 5\) and \(y=2\) into \(7x+3y\).
We have \(7(5)+3(2)=35 + 6=41\).
Since \(41
eq2\), \((5,2)\) is not a solution.

Answer:

• For \((0,-8)\): No
• For \((2,-4)\): Yes
• For \((-1,3)\): Yes
• For \((5,2)\): No