Let’s denote:

• Number of boys = \(x\)
• Number of girls = \(y\)

From the given conditions:

1. Total number of students: \(x + y = 10\)
2. Number of girls is 4 more than boys: \(y = x + 4\)

This forms our pair of linear equations:

To solve graphically, we’ll find some points on each line:

For \(x + y = 10\):

\(x\)	\(y\)	Point
0	10	(0, 10)
5	5	(5, 5)
10	0	(10, 0)

For \(y = x + 4\):

\(x\)	\(y\)	Point
0	4	(0, 4)
3	7	(3, 7)
6	10	(6, 10)

The intersection point is \((3, 7)\), which means:

• Number of boys = 3
• Number of girls = 7

Let’s denote:

• Cost of one pencil = \(x\) rupees
• Cost of one pen = \(y\) rupees

From the given information:

1. 5 pencils and 7 pens cost ₹50: \(5x + 7y = 50\)
2. 7 pencils and 5 pens cost ₹46: \(7x + 5y = 46\)

To solve algebraically, let’s use elimination method:

Multiply equation (1) by 7:

Multiply equation (2) by 5:

Subtract equation (4) from equation (3):

Substitute \(y = 5\) back into equation (1):

To verify graphically, let’s plot the lines:

For \(5x + 7y = 50\):

\(x\)	\(y\)	Point
0	7.14	(0, 7.14)
10	0	(10, 0)

For \(7x + 5y = 46\):

\(x\)	\(y\)	Point
0	9.2	(0, 9.2)
6.57	0	(6.57, 0)

The solution is \((3, 5)\), which means:

• Cost of one pencil = ₹3
• Cost of one pen = ₹5

We need to rewrite both equations in the standard form \(ax + by + c = 0\) and compare the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}\) and \(\frac{c_1}{c_2}\).

Comparing the coefficients:

• \(\frac{a_1}{a_2} = \frac{5}{7}\)
• \(\frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}\)
• \(\frac{c_1}{c_2} = \frac{8}{-9} = \frac{-8}{9}\)

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines intersect at a point.

Rewriting in the standard form \(ax + by + c = 0\):

Comparing the coefficients:

• \(\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}\)
• \(\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}\)
• \(\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}\)

Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines are coincident (same line).

We can verify this by simplifying equation (2):

Which is identical to equation (1).

Rewriting in the standard form \(ax + by + c = 0\):

Let’s multiply equation (2) by 3 to make the comparison clearer:

Now comparing equations (1) and (3):

• \(\frac{a_1}{a_2} = \frac{6}{6} = 1\)
• \(\frac{b_1}{b_2} = \frac{-3}{-3} = 1\)
• \(\frac{c_1}{c_2} = \frac{10}{27} \neq 1\)

Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel.

Rewriting in the standard form \(ax + by + c = 0\):

Comparing the coefficients:

• \(\frac{a_1}{a_2} = \frac{3}{2}\)
• \(\frac{b_1}{b_2} = \frac{2}{-3} = \frac{-2}{3}\)
• \(\frac{c_1}{c_2} = \frac{-5}{-7} = \frac{5}{7}\)

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines intersect at a point, so the system is consistent.

Let’s solve the system to find the solution:

Multiply equation (1) by 2:

Multiply equation (2) by 3:

Subtract equation (4) from equation (3):

Substitute \(y = \frac{-11}{13}\) into equation (1):

Therefore, the solution is \(x = \frac{29}{13}, y = \frac{-11}{13}\).

Rewriting in the standard form \(ax + by + c = 0\):

Let’s simplify equation (2) by dividing by 2:

Now comparing equations (1) and (3):

• \(\frac{a_1}{a_2} = \frac{2}{2} = 1\)
• \(\frac{b_1}{b_2} = \frac{-3}{-3} = 1\)
• \(\frac{c_1}{c_2} = \frac{-8}{-4.5} = \frac{8}{4.5} \neq 1\)

Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel, so the system is inconsistent with no solution.

Let’s convert the first equation to eliminate fractions by multiplying by 6 (LCM of 2 and 3):

Second equation:

Rewriting in the standard form \(ax + by + c = 0\):

Comparing the coefficients:

• \(\frac{a_1}{a_2} = \frac{9}{9} = 1\)
• \(\frac{b_1}{b_2} = \frac{10}{-10} = -1 \neq 1\)

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the lines intersect at a point, so the system is consistent.

Let’s solve the system to find the solution:

Adding equations (1) and (2):

Substitute \(x = \frac{28}{9}\) into equation (1):

Therefore, the solution is \(x = \frac{28}{9}, y = \frac{7}{5}\).

Rewriting in the standard form \(ax + by + c = 0\):

Let’s multiply equation (1) by 2:

Adding equations (2) and (3):

This gives us a true statement, which means the two equations are equivalent.

Comparing the coefficients formally:

• \(\frac{a_1}{a_2} = \frac{5}{-10} = \frac{-1}{2}\)
• \(\frac{b_1}{b_2} = \frac{-3}{6} = \frac{-1}{2}\)
• \(\frac{c_1}{c_2} = \frac{-11}{22} = \frac{-1}{2}\)

Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines are coincident, so the system is consistent with infinitely many solutions.

The solution can be expressed as \(\{(x, y) | 5x – 3y = 11\}\), or equivalently, \(y = \frac{5x – 11}{3}\).

Let’s convert the first equation to eliminate fractions by multiplying by 3:

Second equation:

Let’s multiply equation (2) by 2:

Comparing equations (1) and (3), we can see they are identical.

Formally, rewriting in the standard form \(ax + by + c = 0\):

Comparing the coefficients:

• \(\frac{a_1}{a_2} = \frac{4}{2} = 2\)
• \(\frac{b_1}{b_2} = \frac{6}{3} = 2\)
• \(\frac{c_1}{c_2} = \frac{-24}{-12} = 2\)

Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines are coincident, so the system is consistent with infinitely many solutions.

The solution can be expressed as \(\{(x, y) | 2x + 3y = 12\}\), or equivalently, \(y = \frac{12 – 2x}{3}\).

Let’s simplify the second equation by dividing by 2:

We can see that both equations are identical:

Rewriting in the standard form \(ax + by + c = 0\):

Comparing the coefficients:

• \(\frac{a_1}{a_2} = \frac{1}{1} = 1\)
• \(\frac{b_1}{b_2} = \frac{1}{1} = 1\)
• \(\frac{c_1}{c_2} = \frac{-5}{-5} = 1\)

Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines are coincident, so the system is consistent with infinitely many solutions.

Graphically, we have a single line \(x + y = 5\) representing infinitely many solutions.

The solution can be expressed as \(\{(x, y) | x + y = 5\}\), or equivalently, \(y = 5 – x\).

Let’s simplify the second equation by dividing by 3:

Now we have:

Since the left sides are identical but the right sides are different (\(8 \neq \frac{16}{3}\)), these equations represent parallel lines.

Rewriting in the standard form \(ax + by + c = 0\):

Comparing the coefficients:

• \(\frac{a_1}{a_2} = \frac{1}{1} = 1\)
• \(\frac{b_1}{b_2} = \frac{-1}{-1} = 1\)
• \(\frac{c_1}{c_2} = \frac{-8}{-\frac{16}{3}} = \frac{8}{\frac{16}{3}} = \frac{8 \times 3}{16} = \frac{24}{16} = \frac{3}{2} \neq 1\)

Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel, so the system is inconsistent with no solution.

Graphically, we can see that the lines are parallel with no intersection point, confirming that the system is inconsistent.

Rewriting the equations in slope-intercept form:

Solving the system of equations:

From (1): \(y = 6 – 2x\)

Substituting into (2):

Substituting \(x = 2\) into equation (1):

Therefore, the solution is \((2, 2)\).

To confirm if the system is consistent, let’s check if the point \((2, 2)\) satisfies both equations:

For equation 1: \(2(2) + 2 – 6 = 4 + 2 – 6 = 0\) ✓

For equation 2: \(4(2) – 2(2) – 4 = 8 – 4 – 4 = 0\) ✓

The point satisfies both equations, so the system is consistent with a unique solution.

Graphically, we can see that the lines intersect at the point \((2, 2)\), confirming that the system is consistent with a unique solution.

Simplifying the first equation:

Simplifying the second equation:

Since \(1 \neq \frac{5}{4}\), these equations represent parallel lines with different y-intercepts.

Rewriting in the standard form \(ax + by + c = 0\):

Comparing the coefficients:

• \(\frac{a_1}{a_2} = \frac{1}{4} = \frac{1}{4}\)
• \(\frac{b_1}{b_2} = \frac{-1}{-4} = \frac{1}{4}\)
• \(\frac{c_1}{c_2} = \frac{-1}{-5} = \frac{1}{5} \neq \frac{1}{4}\)

Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the lines are parallel, so the system is inconsistent with no solution.

Graphically, we can see that the lines are parallel with no intersection point, confirming that the system is inconsistent.

Let’s denote the number as x.

Then its reciprocal is \(\frac{1}{x}\).

From the given condition:

Multiplying both sides by x:

Using the quadratic formula:

So, x = \(2 + \sqrt{3}\) or x = \(2 – \sqrt{3}\).

Verification:

Therefore, the numbers are \(2 + \sqrt{3} \approx 3.732\) and \(2 – \sqrt{3} \approx 0.268\).

Let’s denote:

• Speed of the boat in still water = \(x\) km/h
• Speed of the stream = \(y\) km/h

Then:

• Speed of the boat upstream = \((x – y)\) km/h
• Speed of the boat downstream = \((x + y)\) km/h

From the first given condition:

From the second given condition:

Let’s simplify by setting \(u = \frac{1}{x – y}\) and \(v = \frac{1}{x + y}\).

Then our equations become:

Multiplying equation (1) by 4:

Multiplying equation (2) by 3:

Subtracting equation (4) from equation (3):

Substituting back into equation (1):

Now, we have:

Solving these two equations:

Adding equations (5) and (6):

Substituting into equation (5):

Therefore:

• Speed of the boat in still water = 8 km/h
• Speed of the stream = 3 km/h

Given the system of equations:

And the unique solution (2, -1).

Substituting (2, -1) into equation (1):

Substituting (2, -1) into equation (2):

Therefore, p = 1 and q = 1.

Let’s denote:

• Speed of the train = \(x\) km/h
• Speed of the car = \(y\) km/h

From the first condition:

From the second condition (18 minutes = 0.3 hours):

Let’s multiply equation (1) by 13:

Let’s multiply equation (2) by 25:

Subtracting equation (3) from equation (4):

Substituting back into equation (1):

Therefore:

• Speed of the train = 100 km/h
• Speed of the car = 80 km/h

Given line: 3x – 4y + 2 = 0

Let’s rewrite it in slope-intercept form:

So the slope of the given line is \(\frac{3}{4}\).

For a parallel line passing through the point (2, 3), the equation in point-slope form is:

Therefore, the equation of the line parallel to 3x – 4y + 2 = 0 and passing through (2, 3) is: