Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 6 · Chapter companion part
Explain a simultaneous solution, substitute a complete expression and check both original equations, including reciprocal substitutions.
Class 8 English Version: part A of the chapter 6 companion. Eight teaching sections, 30 written tasks and a notebook challenge. Selected companion coverage, not complete textbook coverage; use all four parts. Numeric checks cover only the requested value; full equations, graphs, conditions and explanations need guardian or teacher review.
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A linear equation in x and y has the form ax + by = c, with fixed coefficients a,b not both zero. Each variable has first power and there is no product xy. One equation such as x + y = 9 admits many real ordered pairs. A simultaneous solution must satisfy every equation with the same x and y. The order matters: (6,3) means x=6 and y=3, not the other way round.
For x + y = 9 and x − y = 3, the pair (6,3) gives 9 and 3, so it satisfies both. The pair (5,4) gives the first total but difference 1, so it is not a simultaneous solution. Calculate the left-hand side of each original equation separately. Finding any pair on one line does not finish the problem.
Substitution means replacing a variable by an equal complete expression. From x − y = 3 obtain x = y + 3. In x + y = 9 replace x, giving (y + 3) + y = 9. Thus 2y = 6, y = 3 and x = 6. Substituting back into only the rearranged equation is not a full verification; use both original equations.
If y = 2x − 1 and x + 3y = 18, write x + 3(2x − 1) = 18. Distribution gives 7x − 3 = 18, so x = 3 and y = 5. The expression 3(2x − 1) is 6x − 3, not 6x − 1. Brackets preserve the quantity being replaced and prevent a sign or multiplier from disappearing.
For x/2 + y/3 = 4, multiply every term by 6 to obtain 3x + 2y = 24. Combine it with x − y = 3: substitute x = y + 3 to obtain 5y + 9 = 24, so y = 3 and x = 6. Multiplying only the fractions and leaving the right-hand side at 4 would create a different equation. A fixed nonzero multiplier keeps the solutions unchanged.
For 2/x + 3/y = 2 and 4/x − 3/y = 1, require x,y nonzero. Set u = 1/x and v = 1/y. Then 2u + 3v = 2 and 4u − 3v = 1. Their sum gives u = 1/2; then v = 1/3, so x = 2 and y = 3. Do not report u and v as x and y. If a transformed solution has u=0, no finite nonzero x can have reciprocal zero.
On printed page 102 the source goes from 2y = 4 to a line saying y = 7 − 3, then states y = 2. The valid step is y = 4/2 = 2; 7 − 3 equals 4. On printed pages 103–104, example 3 asks 2y + 5z = 16 and y − 2z = 1, but its solution uses −1 in the second equation. As stated, z = 14/9 and y = 37/9. The changed −1 system gives z = 2 and y = 3. Keep these different systems separate.
Choose an equation where a variable already has coefficient 1 or −1 when possible. Isolate it by doing the same operation on both sides, substitute into the other equation, solve and restore the second variable. Fractions in a valid answer are allowed. A method is justified by preserving equality, not by a rule that a term magically jumps across the sign.
For x=3y+2 and 2x+y=18, replace the whole x by (3y+2): 6y+4+y=18, so y=2,x=8. Both original equations hold.
An automatic check may ask only for x or y. Your written answer still needs the complete pair and both original checks.
The solution of a linear system need not be integral. Keep exact fractions until a task explicitly asks for a rounded graph reading.
The equations are x+y=9 and x−y=3. A shared solution must pass both.
| Candidate (x,y) | x+y | x−y | Shared solution? |
|---|---|---|---|
| (6,3) | 9 | 3 | Yes |
| (5,4) | 9 | 1 | No: difference fails |
| (3,6) | 9 | −3 | No: coordinate order matters |
Matching a single equation is not sufficient. Negative or fractional coordinates may be valid in other systems.
From y=2x−1, replace every y in x+3y=18.
x + 3(2x−1) = 18 7x−3 = 18 x = 3; y = 5
The numeric checker asks for a specified value. These written steps and both checks remain part of the task.
The original x and y must be nonzero.
2/x + 3/y = 2; 4/x − 3/y = 1 u=1/x, v=1/y 6u=3; u=1/2; v=1/3 x=2; y=3
A transformed zero reciprocal does not represent a finite original variable.
Example 3 changes a sign between printed 103 and 104; the original question and altered solution cannot be mixed.
| Version | Equations | (y,z) | Check of second equation |
|---|---|---|---|
| Stated question | 2y+5z=16; y−2z=1 | (37/9,14/9) | 37/9−28/9=1 |
| Changed solution | 2y+5z=16; y−2z=−1 | (3,2) | 3−4=−1 |
The source page 102 arithmetic line y=7−3 also does not follow from 2y=4. The correct division is y=4/2=2.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original reasoning table; cell widths do not represent numerical magnitudes.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original symbolic guide; arrows show reasoning order, not forces or a measured scale.
Show your working on paper. Check your answer, then choose Next.
2
The distinct variables are x and y; 4,7,20 are fixed numbers.
9
6+3=9; the difference condition must be checked separately.
1
5−4=1, so this pair fails the condition x−y=3.
6
Substitute x=y+3: 2y+3=9 gives y=3, x=6.
3
The same solution pair is (6,3), checked in both originals.
-3
Three multiplies −1 as well as 2x.
3
7x−3=18 gives x=3; then y=5.
5
At x=3, y=6−1=5; 3+3×5=18.
24
The multiplier applies to every term, including 4.
3
3(y+3)+2y=24 gives 5y=15.
2
Adding gives 6/x=3, hence x=2; all original denominators are nonzero.
3
With x=2, 1+3/y=2 gives y=3.
2
Divide both sides by the coefficient 2; the answer is y=2.
14
y=2z+1 gives 9z+2=16, hence z=14/9.
2
y=2z−1 gives 9z−2=16, so z=2 and y=3.
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For x+y=9, choose any real x and set y=9−x. For example (0,9), (6,3) and (10,−1) all work. A second independent condition may select one of them.
Give at least two pairs and explain how they arise.
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Both give sum 9, but their differences are 3 and −3 respectively. Only (6,3) satisfies both conditions in the stated order.
Check both left-hand sides for each pair.
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x=y+4, so 2y+4=12, y=4 and x=8. Verify 8+4=12 and 8−4=4.
Show the replacement and both final checks.
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2(3y+2)+y=18 gives 7y=14. Thus y=2,x=8; 2×8+2=18.
The coefficient must multiply both terms in brackets.
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The correct expansion is 10−2y. Two multiplies every term inside the bracket; at y=1 the original is 8, agreeing with 10−2 but not 10−1.
Give a structural reason and a numerical check.
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x=y. Multiply the first equation by 6: 2x+3y=30. Thus 5x=30, x=y=6. Each original is satisfied.
Preserve the right-hand side when clearing denominators.
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Set u=1/x,v=1/y with x,y≠0. Add to get 2u=1, u=1/2; then v=1/4. Hence x=2,y=4. Verify the original fractions.
Restore the reciprocals; do not report (1/2,1/4) as (x,y).
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For a finite real nonzero x, 1/x cannot equal zero. Multiplying by x would give 1=0. Reject this transformed value as outside the reciprocal substitution.
Explain the domain rather than dividing by zero.
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Add to get 2x/a=2/a, so x=1. Subtract to obtain 2y/b=2/b, so y=1. Require a,b nonzero and verify both original equations.
State which quantities are fixed parameters and keep their restrictions.
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The intermediate printed y=7−3 would equal 4 and does not follow from 2y=4. Divide by 2 to obtain y=2; keep the correction beside the source wording.
Distinguish the wrong intermediate step from the correct final value.
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Substitute y=2z+1 into 2y+5z=16: 9z=14, so z=14/9,y=37/9. The pair (3,2) has y−2z=−1, not 1; it solves the changed system.
Check both versions explicitly; do not silently replace a sign.
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Use y=19−4x. Then x−2(19−4x)=−2 gives 9x=36, x=4 and y=3. Check 16+3=19 and 4−6=−2.
A different valid replacement is acceptable with correct working.
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Find y from an original equation, state the ordered pair in the defined order, and substitute that pair into both originals. A single x-coordinate does not itself establish a simultaneous solution.
Use a complete pair and both checks in an example.
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The product xy is not ax+by with fixed coefficients; the coefficient of x would depend on y. First powers written separately are not sufficient when variables multiply each other.
Contrast it with a valid linear example, such as 2x+3y=6.
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One choice is x+y=3 and 2x−y=−9. Adding gives 3x=−6, x=−2 and y=5. Many other independently constraining pairs are valid.
Check the same pair in both equations and explain whether the solution is unique.
Make a two-condition solution card and a reciprocal-substitution card.
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This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 6, printed pages 101–120. It does not replace the complete chapter or its practical teaching.
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