GSEB Class 10 Maths Solutions Chapter 3 દ્વિચલ સુરેખ સમીકરણયુગ્મ Exercise 3.3

Question 1. નીચેનાં દ્વિચલ સુરેખ સમીકરણયુગ્મનો ઉકેલ આદેશની રીતે મેળવોઃ
(i) \( x + y = 14 \)
\( x - y = 4 \)
(ii) \( s - t = 3 \)
\( \frac{s}{3} + \frac{t}{2} = 6 \)
(iii) \( 3x - y = 3 \)
\( 9x - 3y = 9 \)
(iv) \( 0.2x + 0.3y = 1.3 \)
\( 0.4x + 0.5y = 2.3 \)
(v) \( \sqrt{2}x + \sqrt{3}y = 0 \)
\( \sqrt{3}x - \sqrt{8}y = 0 \)
(vi) \( \frac{3x}{2} - \frac{5y}{3} = -2 \)
\( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \)
Answer:
(i) Given equations are:
\( x + y = 14 \) ..........(1)
\( x - y = 4 \) ..........(2)
From equation (1), we get \( y = 14 - x \).
Substitute this value of y into equation (2):
\( x - (14 - x) = 4 \)
\( x - 14 + x = 4 \)
\( 2x - 14 = 4 \)
\( 2x = 4 + 14 \)
\( 2x = 18 \)
\( x = 9 \)
Now, substitute \( x = 9 \) into equation (1):
\( 9 + y = 14 \)
\( y = 14 - 9 \)
\( y = 5 \)
Thus, the solution for the given pair of linear equations is \( x = 9, y = 5 \).
Verification:
\( x + y = 9 + 5 = 14 \)
\( x - y = 9 - 5 = 4 \)
The solution satisfies both given equations.
In simple words: We found \( y \) from the first equation and put it into the second one to get \( x \). Then, we used that \( x \) value to find \( y \). The solution is \( x=9 \) and \( y=5 \).

Exam Tip: In the substitution method, always isolate one variable in one equation and substitute its expression into the other equation. Then, remember to verify your answer using both original equations.

Answer:
(ii) Given equations are:
\( s - t = 3 \) ..........(1)
\( \frac{s}{3} + \frac{t}{2} = 6 \) ..........(2)
From equation (1), we get \( s = t + 3 \).
Substitute this value of s into equation (2):
\( \frac{t+3}{3} + \frac{t}{2} = 6 \)
To remove the denominators, multiply the entire equation by the Least Common Multiple (LCM) of 3 and 2, which is 6:
\( 6 \times \left( \frac{t+3}{3} \right) + 6 \times \left( \frac{t}{2} \right) = 6 \times 6 \)
\( 2(t+3) + 3t = 36 \)
\( 2t + 6 + 3t = 36 \)
\( 5t + 6 = 36 \)
\( 5t = 36 - 6 \)
\( 5t = 30 \)
\( t = 6 \)
Now, substitute \( t = 6 \) into equation (1):
\( s - 6 = 3 \)
\( s = 3 + 6 \)
\( s = 9 \)
Thus, the solution for the given pair of linear equations is \( s = 9, t = 6 \).
Verification:
\( s - t = 9 - 6 = 3 \)
\( \frac{s}{3} + \frac{t}{2} = \frac{9}{3} + \frac{6}{2} = 3 + 3 = 6 \)
The solution satisfies both given equations.
In simple words: We changed the first equation to get 's' by itself. Then we put that 's' into the second equation. After simplifying the fractions, we found 't'. Using 't', we then found 's'. The answer is \( s=9 \) and \( t=6 \).

Exam Tip: When dealing with fractional equations, always simplify them first by multiplying by the LCM of the denominators. This makes the substitution process much easier and reduces errors.

Answer:
(iii) Given equations are:
\( 3x - y = 3 \) ..........(1)
\( 9x - 3y = 9 \) ..........(2)
From equation (1), we can express y as: \( y = 3x - 3 \).
Substitute this value of y into equation (2):
\( 9x - 3(3x - 3) = 9 \)
\( 9x - 9x + 9 = 9 \)
\( 9 = 9 \)
Here, we get a true statement, \( 9 = 9 \), which does not give any specific value for x or y. This indicates that the given pair of linear equations has infinitely many solutions. This happens because the two equations are dependent (one is a multiple of the other).
Any pair \( (x, y) \) that satisfies \( y = 3x - 3 \), where x is any real number, will be a solution to the system.
In simple words: When we tried to solve these equations, we ended up with 9 = 9. This means the two equations are basically the same, just written differently. So, there are many possible answers, and 'y' is always 3 times 'x' minus 3.

Exam Tip: If, after substitution, you arrive at a true statement (like 9=9), it means the system has infinitely many solutions, implying the lines are coincident. If you get a false statement (like 0=5), it means there are no solutions, and the lines are parallel.

Answer:
(iv) Given equations are:
\( 0.2x + 0.3y = 1.3 \) ..........(1)
\( 0.4x + 0.5y = 2.3 \) ..........(2)
To make calculations simpler, multiply both equations by 10 to clear the decimals:
From (1): \( 10 \times (0.2x + 0.3y) = 10 \times 1.3 \implies 2x + 3y = 13 \) ..........(3)
From (2): \( 10 \times (0.4x + 0.5y) = 10 \times 2.3 \implies 4x + 5y = 23 \) ..........(4)
From equation (3), express x in terms of y:
\( 2x = 13 - 3y \)
\( x = \frac{13 - 3y}{2} \)
Substitute this value of x into equation (4):
\( 4 \left( \frac{13 - 3y}{2} \right) + 5y = 23 \)
\( 2(13 - 3y) + 5y = 23 \)
\( 26 - 6y + 5y = 23 \)
\( 26 - y = 23 \)
\( -y = 23 - 26 \)
\( -y = -3 \)
\( y = 3 \)
Now, substitute \( y = 3 \) back into the expression for x:
\( x = \frac{13 - 3(3)}{2} \)
\( x = \frac{13 - 9}{2} \)
\( x = \frac{4}{2} \)
\( x = 2 \)
Thus, the solution for the given pair of linear equations is \( x = 2, y = 3 \).
In simple words: We got rid of the decimal points by multiplying both equations by 10. Then, we solved for \( x \) from the first equation and put that into the second equation to find \( y \). Finally, we used \( y \) to find \( x \). The answer is \( x=2 \) and \( y=3 \).

Exam Tip: For equations with decimals, it is often easier to first multiply by a power of 10 to convert them into integer coefficients. This minimizes calculation errors during substitution.

Answer:
(v) Given equations are:
\( \sqrt{2}x + \sqrt{3}y = 0 \) ..........(1)
\( \sqrt{3}x - \sqrt{8}y = 0 \) ..........(2)
From equation (1), express x in terms of y:
\( \sqrt{2}x = -\sqrt{3}y \)
\( x = -\frac{\sqrt{3}}{\sqrt{2}}y \)
Substitute this value of x into equation (2):
\( \sqrt{3} \left( -\frac{\sqrt{3}}{\sqrt{2}}y \right) - \sqrt{8}y = 0 \)
\( -\frac{3}{\sqrt{2}}y - \sqrt{8}y = 0 \)
We know \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \). So the equation becomes:
\( -\frac{3}{\sqrt{2}}y - 2\sqrt{2}y = 0 \)
Multiply the entire equation by \( \sqrt{2} \) to clear the denominator:
\( -3y - 2\sqrt{2} \times \sqrt{2}y = 0 \)
\( -3y - 4y = 0 \)
\( -7y = 0 \)
\( y = 0 \)
Now, substitute \( y = 0 \) back into the expression for x:
\( x = -\frac{\sqrt{3}}{\sqrt{2}}(0) \)
\( x = 0 \)
Thus, the solution for the given pair of linear equations is \( x = 0, y = 0 \).
In simple words: We solved the first equation for \( x \) in terms of \( y \). Then we put that into the second equation. We simplified the square roots and solved for \( y \), which turned out to be 0. When \( y \) is 0, \( x \) is also 0. So, both \( x \) and \( y \) are 0.

Exam Tip: When solving equations with square roots, try to simplify the roots first (e.g., \( \sqrt{8} = 2\sqrt{2} \)). This makes the algebraic manipulation easier and helps avoid common calculation mistakes.

Answer:
(vi) Given equations are:
\( \frac{3x}{2} - \frac{5y}{3} = -2 \) ..........(1)
\( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \) ..........(2)
First, we simplify equation (1) by multiplying by the LCM of 2 and 3, which is 6:
\( 6 \times \left( \frac{3x}{2} \right) - 6 \times \left( \frac{5y}{3} \right) = 6 \times (-2) \)
\( 9x - 10y = -12 \) ..........(3)
Next, we simplify equation (2) by multiplying by the LCM of 3, 2, and 6, which is 6:
\( 6 \times \left( \frac{x}{3} \right) + 6 \times \left( \frac{y}{2} \right) = 6 \times \left( \frac{13}{6} \right) \)
\( 2x + 3y = 13 \) ..........(4)
From equation (3), express x in terms of y:
\( 9x = 10y - 12 \)
\( x = \frac{10y - 12}{9} \)
Substitute this value of x into equation (4):
\( 2 \left( \frac{10y - 12}{9} \right) + 3y = 13 \)
Multiply the entire equation by 9 to clear the denominator:
\( 2(10y - 12) + 9(3y) = 9(13) \)
\( 20y - 24 + 27y = 117 \)
\( 47y - 24 = 117 \)
\( 47y = 117 + 24 \)
\( 47y = 141 \)
\( y = \frac{141}{47} \)
\( y = 3 \)
Now, substitute \( y = 3 \) back into the expression for x:
\( x = \frac{10(3) - 12}{9} \)
\( x = \frac{30 - 12}{9} \)
\( x = \frac{18}{9} \)
\( x = 2 \)
Thus, the solution for the given pair of linear equations is \( x = 2, y = 3 \).
In simple words: First, we made both equations simpler by removing the fractions. Then, we solved for \( x \) from the first new equation and put that into the second new equation. This helped us find \( y \). Finally, we used \( y \) to get \( x \). The answer is \( x=2 \) and \( y=3 \).

Exam Tip: Always clear fractions by multiplying by the LCM of denominators before attempting substitution. This prevents fractional arithmetic errors in the subsequent steps.

Question 2. 2x + y = 11 અને 2x – 4y = -24નો ઉકેલ શોધો અને એવો 'm' શોધો કે જેથી y = mx + 3 થાય.
Answer:
Given linear equations are:
\( 2x + y = 11 \) ..........(1)
\( 2x - 4y = -24 \) ..........(2)
From equation (1), we can express y as: \( y = 11 - 2x \).
Substitute this value of y into equation (2):
\( 2x - 4(11 - 2x) = -24 \)
\( 2x - 44 + 8x = -24 \)
\( 10x - 44 = -24 \)
\( 10x = -24 + 44 \)
\( 10x = 20 \)
\( x = 2 \)
Now, substitute \( x = 2 \) back into the expression for y:
\( y = 11 - 2(2) \)
\( y = 11 - 4 \)
\( y = 7 \)
So, the solution to the system of equations is \( x = 2, y = 7 \).
Next, we need to find the value of 'm' such that \( y = mx + 3 \).
Substitute the values of \( x = 2 \) and \( y = 7 \) into this equation:
\( 7 = m(2) + 3 \)
\( 7 = 2m + 3 \)
\( 7 - 3 = 2m \)
\( 4 = 2m \)
\( m = \frac{4}{2} \)
\( m = 2 \)
Thus, the solution to the system is \( x = 2, y = 7 \), and the value of 'm' is 2.
In simple words: First, we found \( x=2 \) and \( y=7 \) by solving the two equations. Then, we used these \( x \) and \( y \) values in the equation \( y = mx + 3 \) to find that \( m \) is 2.

Exam Tip: When a question asks for both the solution to a system of equations and an additional variable (like 'm'), solve the system first to get the values of 'x' and 'y', and then use those values to find the additional variable.

Question 3. નીચેની સમસ્યા ઉપરથી દ્વિચલ સુરેખ સમીકરણયુગ્મ મેળવો અને તેમનો ઉકેલ આદેશની રીતે મેળવોઃ
(i) બે સંખ્યાનો તફાવત 26 છે અને એક સંખ્યા બીજી સંખ્યાથી ત્રણ ગણી છે, તો તે બે સંખ્યા શોધો.
Answer:
Let the larger number be \( x \) and the smaller number be \( y \).
According to the first condition, the difference between the two numbers is 26:
\( x - y = 26 \) ..........(1)
According to the second condition, one number is three times the other. Since x is the larger number, it must be three times the smaller number:
\( x = 3y \) ..........(2)
Substitute the value of \( x \) from equation (2) into equation (1):
\( 3y - y = 26 \)
\( 2y = 26 \)
\( y = \frac{26}{2} \)
\( y = 13 \)
Now, substitute \( y = 13 \) back into equation (2) to find \( x \):
\( x = 3(13) \)
\( x = 39 \)
Thus, the two numbers are 39 and 13.
Verification:
Difference: \( 39 - 13 = 26 \)
One number is three times the other: \( 39 = 3 \times 13 \).
Both conditions are satisfied.
In simple words: We named the two numbers \( x \) and \( y \). We wrote down two equations based on the problem. We found that \( y \) is 13 and \( x \) is 39.

Exam Tip: For word problems, always clearly define your variables (e.g., let 'x' be the larger number, 'y' be the smaller). Formulate the equations carefully based on the problem's conditions before solving.

Answer:
(ii) બે પૂરકકોણો પૈકી મોટો ખૂણો નાના ખૂણા કરતાં 18° મોટો હોય, તો તે પૂરકકોણો શોધો.
Answer:
Let the larger supplementary angle be \( x \) degrees and the smaller supplementary angle be \( y \) degrees.
Since the angles are supplementary, their sum is 180 degrees:
\( x + y = 180 \) ..........(1)
According to the second condition, the larger angle is 18 degrees more than the smaller angle:
\( x = y + 18 \) ..........(2)
Substitute the value of \( x \) from equation (2) into equation (1):
\( (y + 18) + y = 180 \)
\( 2y + 18 = 180 \)
\( 2y = 180 - 18 \)
\( 2y = 162 \)
\( y = \frac{162}{2} \)
\( y = 81 \)
Now, substitute \( y = 81 \) back into equation (2) to find \( x \):
\( x = 81 + 18 \)
\( x = 99 \)
Thus, the two supplementary angles are 99° and 81°.
Verification:
Difference: \( 99^\circ - 81^\circ = 18^\circ \)
Sum: \( 99^\circ + 81^\circ = 180^\circ \), confirming they are supplementary.
In simple words: We assumed the angles were \( x \) and \( y \). We knew they add up to 180 degrees and that one is 18 degrees bigger than the other. We used these facts to make equations and found the angles are 99 degrees and 81 degrees.

Exam Tip: Remember the definitions: complementary angles sum to 90°, and supplementary angles sum to 180°. Clearly setting up these initial equations is crucial for solving word problems involving angles.

Answer:
(iii) ક્રિકેટ-ટીમના કોચે 7 બૅટ અને 6 દડા 3800માં ખરીધા. પછીથી તેણે તે જ કિંમતવાળા ૩ બૅટ અને 5 દડા 1750માં ખરીધા. તો એક બૅટની કિંમત અને એક દડાની કિંમત શોધો.
Answer:
Let the cost of one bat be \( x \) Rs. and the cost of one ball be \( y \) Rs.
According to the first condition, the coach bought 7 bats and 6 balls for Rs. 3800:
\( 7x + 6y = 3800 \) ..........(1)
According to the second condition, he later bought 3 bats and 5 balls for Rs. 1750:
\( 3x + 5y = 1750 \) ..........(2)
From equation (2), express x in terms of y:
\( 3x = 1750 - 5y \)
\( x = \frac{1750 - 5y}{3} \)
Substitute this value of x into equation (1):
\( 7 \left( \frac{1750 - 5y}{3} \right) + 6y = 3800 \)
Multiply the entire equation by 3 to clear the denominator:
\( 7(1750 - 5y) + 3(6y) = 3(3800) \)
\( 12250 - 35y + 18y = 11400 \)
\( 12250 - 17y = 11400 \)
\( -17y = 11400 - 12250 \)
\( -17y = -850 \)
\( y = \frac{-850}{-17} \)
\( y = 50 \)
Now, substitute \( y = 50 \) back into the expression for x:
\( x = \frac{1750 - 5(50)}{3} \)
\( x = \frac{1750 - 250}{3} \)
\( x = \frac{1500}{3} \)
\( x = 500 \)
Thus, the cost of one bat is Rs. 500 and the cost of one ball is Rs. 50.
Verification:
For the first purchase: \( 7(500) + 6(50) = 3500 + 300 = 3800 \). Correct.
For the second purchase: \( 3(500) + 5(50) = 1500 + 250 = 1750 \). Correct.
In simple words: We let \( x \) be the cost of a bat and \( y \) be the cost of a ball. We wrote two equations from the given information about buying bats and balls. After solving, we found that a bat costs Rs. 500 and a ball costs Rs. 50.

Exam Tip: Clearly defining variables and forming the correct equations from word problems is the most important step. Always perform a quick check with the original problem conditions to ensure your answers are reasonable.

Answer:
(iv) એક શહેરમાં ટેક્સીનું ભાડું નિશ્ચિત ભાડા અને અંતરના પ્રમાણમાં સંયુક્ત રીતે લેવાય છે. 10 કિમીના અંતર માટે ₹ 105 અને 15 કિમીની મુસાફરી માટે ₹ 155ની ચુકવણી કરવી પડે છે, તો નિશ્ચિત ભાડું કેટલું અને પ્રતિ કિમી કેટલા દરે કિંમત ચૂકવવી પડે? મુસાફરે 25 કિમીની મુસાફરી માટે કેટલું ભાડું ચૂકવવું પડશે?
Answer:
Let the fixed charge be Rs. \( x \) and the charge per kilometer be Rs. \( y \).
According to the first condition, for a 10 km journey, the fare is Rs. 105:
\( x + 10y = 105 \) ..........(1)
According to the second condition, for a 15 km journey, the fare is Rs. 155:
\( x + 15y = 155 \) ..........(2)
From equation (1), express x in terms of y:
\( x = 105 - 10y \)
Substitute this value of x into equation (2):
\( (105 - 10y) + 15y = 155 \)
\( 105 + 5y = 155 \)
\( 5y = 155 - 105 \)
\( 5y = 50 \)
\( y = \frac{50}{5} \)
\( y = 10 \)
Now, substitute \( y = 10 \) back into the expression for x:
\( x = 105 - 10(10) \)
\( x = 105 - 100 \)
\( x = 5 \)
So, the fixed charge is Rs. 5 and the charge per kilometer is Rs. 10.
Now, we need to find the total fare for a journey of 25 km.
Total fare = Fixed charge + (Charge per km \( \times \) Number of km)
Total fare for 25 km = \( x + 25y \)
Total fare for 25 km = \( 5 + 25(10) \)
Total fare for 25 km = \( 5 + 250 \)
Total fare for 25 km = Rs. 255.
In simple words: We called the fixed charge \( x \) and the per-kilometer charge \( y \). We used the given fares for 10 km and 15 km to make two equations. Solving them, we found the fixed charge is Rs. 5 and the charge per km is Rs. 10. Then we used these to calculate the total fare for 25 km, which is Rs. 255.

Exam Tip: In problems involving fixed charges and variable rates, set up equations for the total cost as "Fixed Charge + (Rate \( \times \) Quantity)". Ensure all parts of the question, including additional calculations, are addressed.

Answer:
(v) એક અપૂર્ણાકના અંશ અને છેદ બંનેમાં 2 ઉમેરતાં તે \(\frac{9}{11}\) બને છે. જો અપૂર્ણાકના અંશ અને છેદ બંનેમાં 3 ઉમેરતાં તે \(\frac{5}{6}\) બને, તો તે અપૂર્ણાંક શોધો.
Answer:
Let the numerator of the fraction be \( x \) and the denominator be \( y \).
So, the fraction is \( \frac{x}{y} \).
According to the first condition, if 2 is added to both the numerator and denominator, the fraction becomes \( \frac{9}{11} \):
\( \frac{x+2}{y+2} = \frac{9}{11} \)
Cross-multiply to get a linear equation:
\( 11(x+2) = 9(y+2) \)
\( 11x + 22 = 9y + 18 \)
\( 11x - 9y = 18 - 22 \)
\( 11x - 9y = -4 \) ..........(1)
According to the second condition, if 3 is added to both the numerator and denominator, the fraction becomes \( \frac{5}{6} \):
\( \frac{x+3}{y+3} = \frac{5}{6} \)
Cross-multiply to get another linear equation:
\( 6(x+3) = 5(y+3) \)
\( 6x + 18 = 5y + 15 \)
\( 6x - 5y = 15 - 18 \)
\( 6x - 5y = -3 \) ..........(2)
Now we have a system of two linear equations:
1) \( 11x - 9y = -4 \)
2) \( 6x - 5y = -3 \)
From equation (2), express x in terms of y:
\( 6x = 5y - 3 \)
\( x = \frac{5y - 3}{6} \)
Substitute this value of x into equation (1):
\( 11 \left( \frac{5y - 3}{6} \right) - 9y = -4 \)
Multiply the entire equation by 6 to clear the denominator:
\( 11(5y - 3) - 6(9y) = 6(-4) \)
\( 55y - 33 - 54y = -24 \)
\( y - 33 = -24 \)
\( y = -24 + 33 \)
\( y = 9 \)
Now, substitute \( y = 9 \) back into the expression for x:
\( x = \frac{5(9) - 3}{6} \)
\( x = \frac{45 - 3}{6} \)
\( x = \frac{42}{6} \)
\( x = 7 \)
Thus, the numerator is 7 and the denominator is 9, so the fraction is \( \frac{7}{9} \).
Verification:
Adding 2: \( \frac{7+2}{9+2} = \frac{9}{11} \). Correct.
Adding 3: \( \frac{7+3}{9+3} = \frac{10}{12} = \frac{5}{6} \). Correct.
In simple words: We let the fraction be \( x/y \). We set up two equations based on what happens when you add 2 or 3 to both the top and bottom of the fraction. After solving these, we found the fraction is \( 7/9 \).

Exam Tip: For fraction-based word problems, always start by defining the fraction as \( \frac{x}{y} \). Convert the word statements into equations by cross-multiplication, then solve the resulting linear system.

Answer:
(vi) પાંચ વર્ષ પછી જેકબની ઉંમર (વર્ષમાં) તેના પુત્રની ઉંમર (વર્ષમાં) કરતાં ત્રણ ગણી હશે. પાંચ વર્ષ પહેલા, જેકબની ઉંમર (વર્ષમાં) તેના પુત્રની ઉંમરથી સાત ગણી હોય, તો તેમની વર્તમાન ઉંમર શોધો.
Answer:
Let Jacob's current age be \( x \) years and his son's current age be \( y \) years.
**Condition 1: Five years after today**
Jacob's age: \( x+5 \) years
Son's age: \( y+5 \) years
Jacob's age will be three times his son's age:
\( x+5 = 3(y+5) \)
\( x+5 = 3y+15 \)
\( x - 3y = 15 - 5 \)
\( x - 3y = 10 \) ..........(1)
**Condition 2: Five years before today**
Jacob's age: \( x-5 \) years
Son's age: \( y-5 \) years
Jacob's age was seven times his son's age:
\( x-5 = 7(y-5) \)
\( x-5 = 7y-35 \)
\( x - 7y = -35 + 5 \)
\( x - 7y = -30 \) ..........(2)
Now we have a system of two linear equations:
1) \( x - 3y = 10 \)
2) \( x - 7y = -30 \)
From equation (1), express x in terms of y:
\( x = 10 + 3y \)
Substitute this value of x into equation (2):
\( (10 + 3y) - 7y = -30 \)
\( 10 - 4y = -30 \)
\( -4y = -30 - 10 \)
\( -4y = -40 \)
\( y = \frac{-40}{-4} \)
\( y = 10 \)
Now, substitute \( y = 10 \) back into the expression for x:
\( x = 10 + 3(10) \)
\( x = 10 + 30 \)
\( x = 40 \)
Thus, Jacob's current age is 40 years and his son's current age is 10 years.
Verification:
Five years later: Jacob = 45, Son = 15. \( 45 = 3 \times 15 \). Correct.
Five years ago: Jacob = 35, Son = 5. \( 35 = 7 \times 5 \). Correct.
In simple words: We used \( x \) for Jacob's age and \( y \) for his son's age. We set up equations for their ages five years in the future and five years in the past. After solving these equations, we found Jacob is 40 years old and his son is 10 years old.

Exam Tip: When dealing with age problems, clearly define current ages. Then, write expressions for ages in the past or future. This helps prevent errors when forming the equations. Always double-check your final ages against the given conditions.