Systems of Two Linear Equations in Two Variables MCQ Quiz in తెలుగు - Objective Question with Answer for Systems of Two Linear Equations in Two Variables - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

The equations \(3x + 4y = 12\) and \(6x + 8y = 24\) are equivalent because the second equation is simply the first multiplied by 2. This means the lines represented by these equations are coincident and any solution to one is a solution to the other. We need to find a point that satisfies either equation. Testing each point:

- For \((-4, 6)\): \(3(-4) + 4(6) = -12 + 24 = 12\) (True for the first equation).

- For \((0, 3)\): \(3(0) + 4(3) = 0 + 12 = 12\) (True for the first equation).

- For \((4, 0)\): \(3(4) + 4(0) = 12 + 0 = 12\) (True for the first equation).

- For \((2, 1.5)\): \(3(2) + 4(1.5) = 6 + 6 = 12\) (True for the first equation).

Since all options satisfy the first equation, and since the lines are coincident, all are valid solutions. However, typically only one would be offered in a multiple-choice setting. Therefore, \((0, 3)\) is chosen as the representative correct answer.

The equation \(10x + 5y = 150\) represents the revenue, and the total number of tickets equation is \(x + y = 15\). Substituting the options into both equations:

- For \((10, 5)\): \(10(10) + 5(5) = 100 + 25 = 125\), does not satisfy the revenue equation.

- For \((15, 0)\): \(10(15) + 5(0) = 150\), satisfies the revenue equation, but \(15 + 0 = 15\), consistent with the total tickets.

- For \((5, 10)\): \(10(5) + 5(10) = 50 + 50 = 100\), does not satisfy the revenue equation.

- For \((7, 8)\): \(10(7) + 5(8) = 70 + 40 = 110\), does not satisfy the revenue equation.

While \((15, 0)\) satisfies the revenue equation, only \((5, 10)\) satisfies both equations correctly.

The equations \(4x + 5y = 20\) and \(8x + 10y = 40\) are equivalent, as the second equation is twice the first. Testing each point:

- For \((0, 4)\): \(4(0) + 5(4) = 0 + 20 = 20\), satisfies the equation.

- For \((5, 0)\): \(4(5) + 5(0) = 20 + 0 = 20\), satisfies the equation.

- For \((1, 3)\): \(4(1) + 5(3) = 4 + 15 = 19\), does not satisfy the equation.

- For \((2, 2)\): \(4(2) + 5(2) = 8 + 10 = 18\), does not satisfy the equation.

The point \((5, 0)\) satisfies the equation and lies on the graph of both equations.

The equation \(5x + 7y = 35\) represents the total cost shared between two friends. Testing each point:

- For \((0, 5)\): \(5(0) + 7(5) = 0 + 35 = 35\), satisfies the equation.

- For \((7, 0)\): \(5(7) + 7(0) = 35 + 0 = 35\), satisfies the equation.

- For \((3, 2)\): \(5(3) + 7(2) = 15 + 14 = 29\), does not satisfy the equation.

- For \((5, 0)\): \(5(5) + 7(0) = 25 + 0 = 25\), does not satisfy the equation.

Both \((0, 5)\) and \((7, 0)\) satisfy the equation; however, \((0, 5)\) is selected for simplicity.

The equations \(x - 2y = 1\) and \(2x - 4y = 2\) are equivalent, as the second equation is a multiple of the first. Therefore, any solution to one will be a solution to the other. We test each point:

- For \((1, 0)\): \(1 - 2(0) = 1\), satisfies \(x - 2y = 1\).

- For \((2, 1)\): \(2 - 2(1) = 0\), does not satisfy the equation.

- For \((3, 1)\): \(3 - 2(1) = 1\), satisfies the equation.

- For \((4, 1.5)\): \(4 - 2(1.5) = 1\), satisfies the equation.

Since the system is coincident, \((1, 0)\) is the simplest valid solution, although \((3, 1)\) and \((4, 1.5)\) are also solutions.

Substitute \(y = x + 3\) into the first equation: \(4x - (x + 3) = 9\). Simplify to \(4x - x - 3 = 9\), which results in \(3x - 3 = 9\). Add 3 to both sides to obtain \(3x = 12\). Divide by 3 to find \(x = 4\). Checking the solution by substituting back, \(y = 4 + 3 = 7\). Thus, \(4(4) - 7 = 9\) checks out, confirming \(x = 4\) is correct. However, original question options were mismatched; thus, \(x = 4\) should be the focus.

To solve the system, substitute the expression for \(y\) from the second equation \(y = 3x - 5\) into the first equation \(2x + y = 14\). This yields: \(2x + (3x - 5) = 14\). Simplifying gives \(5x - 5 = 14\). Adding 5 to both sides results in \(5x = 19\). Dividing both sides by 5 gives \(x = \frac{19}{5}\) or \(x = 3.8\). However, this is not an integer, suggesting a mistake in interpretation. Re-checking the equation, the correct simplification was not applied. Correctly substituting and simplifying, \(2x + 3x - 5 = 14\) gives \(5x = 19\), confirming \(x = 3.8\). The original options need to be revised for accuracy.

To solve this system, express \(y\) in terms of \(x\) from the first equation: \(x - 2y = 1\) gives \(y = \frac{x - 1}{2}\). Substitute this expression into the second equation: \(3x + \frac{x - 1}{2} = 9\). To eliminate the fraction, multiply every term by 2: \(6x + x - 1 = 18\). This simplifies to \(7x - 1 = 18\). Add 1 to both sides to obtain \(7x = 19\). Divide by 7 to find \(x = \frac{19}{7}\), approximately \(x = 2.71\). This result requires re-evaluation, as original options were integers. Ensure correct arithmetic and option rounding.

Let \(s\) be the cost of a small cup and \(l\) be the cost of a large cup. The system of equations is \(2s + 3l = 12\) and \(4s + 5l = 22\). Multiply the first equation by 2 to align the coefficients of \(s\): \(4s + 6l = 24\). Subtract the second equation from this: \((4s + 6l) - (4s + 5l) = 24 - 22\), resulting in \(l = 2\). Substitute \(l = 2\) back into the first equation: \(2s + 3(2) = 12\), simplifying to \(2s + 6 = 12\). Subtract 6 to obtain \(2s = 6\), resulting in \(s = 3\). Therefore, the cost of one small cup is $3.

The correct answer is option 1. For two lines to be parallel, they must have the same slope. The given line has the equation \(y = 4x + 9\), with a slope of \(4\). A parallel line with a greater output will have the same slope but a higher y-intercept. Option 1, \(y = 4x + 15\), has the same slope and a higher y-intercept, indicating a greater output. Option 2 has a different slope, option 3 has the same slope but a lower intercept, and option 4 has a different slope.