Question

which statement describes the graph of the given system of equations?
1.5x + 0.2y = 2.68
1.6 + 0.3y = 2.98

the lines overlap at all
points.

the lines intersect at
(3.1, 0.5).

the lines are parallel.

the lines intersect at
(1.6, 1.4).

Explanation:

Step1: Solve the second equation for y

The second equation is \(1.6 + 0.3y = 2.98\). Subtract 1.6 from both sides:
\(0.3y = 2.98 - 1.6\)
\(0.3y = 1.38\)
Divide both sides by 0.3:
\(y=\frac{1.38}{0.3}=4.6\)? Wait, no, wait, maybe I made a mistake. Wait, 2.98 - 1.6 is 1.38? Wait, 1.6 + 1.38 is 2.98? Wait, 1.6 + 1.38 = 2.98? 1.6 + 1 = 2.6, 2.6 + 0.38 = 2.98, yes. Then 1.38 / 0.3: 0.34 = 1.2, 1.38 - 1.2 = 0.18, 0.18/0.3 = 0.6, so y = 4.6? Wait, that can't be right. Wait, maybe the second equation is \(1.6x + 0.3y = 2.98\)? Oh! Maybe a typo in the original problem. Let's assume that the second equation is \(1.6x + 0.3y = 2.98\) (since otherwise the second equation is a vertical line or something, but the first is a line with x and y). Let's proceed with that correction (since otherwise the second equation is \(0.3y = 1.38\), y=4.6, and the first equation: 1.5x + 0.24.6 = 2.68 → 1.5x + 0.92 = 2.68 → 1.5x = 1.76 → x≈1.17, which is not any of the options. So likely a typo, the second equation is \(1.6x + 0.3y = 2.98\)).

So let's solve the system:

Equation 1: \(1.5x + 0.2y = 2.68\)

Equation 2: \(1.6x + 0.3y = 2.98\)

Let's use elimination. Multiply Equation 1 by 3: \(4.5x + 0.6y = 8.04\)

Multiply Equation 2 by 2: \(3.2x + 0.6y = 5.96\)

Subtract Equation 2 (multiplied by 2) from Equation 1 (multiplied by 3):

\((4.5x - 3.2x) + (0.6y - 0.6y) = 8.04 - 5.96\)

\(1.3x = 2.08\)

\(x = \frac{2.08}{1.3} = 1.6\)

Now plug x = 1.6 into Equation 1:

\(1.5(1.6) + 0.2y = 2.68\)

\(2.4 + 0.2y = 2.68\)

\(0.2y = 2.68 - 2.4 = 0.28\)

\(y = \frac{0.28}{0.2} = 1.4\)

So the solution is (1.6, 1.4), which is one of the options.

Step2: Check the options

Now, let's check the options. The lines intersect at (1.6, 1.4), so that's the correct statement.

Answer:

The lines intersect at (1.6, 1.4).