Solve the system:
x + y = 7
x − y = 1
Add the equations. The y-terms cancel:
2x = 8
x = 4
Substitute x = 4:
4 + y = 7
y = 3
Both equations are true at (4, 3), so that is the solution.

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Solve the system:
x + y = 7
x − y = 1
Add the equations. The y-terms cancel:
2x = 8
x = 4
Substitute x = 4:
4 + y = 7
y = 3
Both equations are true at (4, 3), so that is the solution.
Where can two equations be true at once?
For y = x + 1 and y = −x + 5, set the expressions equal:
x + 1 = −x + 5
x = 2
y = 3
Both equations are true at (2, 3), so that intersection is the system’s solution.
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Can you free x from the parentheses?
3(x + 2) = 18
3x + 6 = 18
3x = 12
x = 4
Distribute, subtract 6, then divide by 3. Check: 3(4 + 2) = 18.
A step-by-step Grade 8 lesson aligned with 8.EE.7b.
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How can every value of x be correct?
x + 3 = x + 3
Subtract x from both sides and 3 = 3 remains. Because this statement is always true, every real number satisfies the original equation: infinitely many solutions.
Grade 8 standard 8.EE.7a.
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When does an equation have no solution?
x + 2 = x + 5 simplifies to 2 = 5 after subtracting x from both sides. Because the result is a false statement, no value of x can satisfy the original equation.
A cinematic Grade 8 lesson aligned with 8.EE.7a.
Can you solve x + 3 = 7 before PuzzleVAI’s drone reaches the answer?
Subtract 3 from both sides. Equality is preserved, the variable is isolated, and x = 4.
A cinematic Grade 8 lesson on inverse operations and one-step linear equations aligned with 8.EE.C.7.
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Chemnitz University of Technology is hiring:
Postdoc in Algebra
Plenary — Simultaneous Equations #maths #mathsteacher #UKMathsChat #GCSEMaths #Plenary #MathsDiscussion #ProblemSolving #Reasoning #SimultaneousEquations #Algebra
#p5js version of "Have a donut" by Andy Sloane from 2006.
www.a1k0n.net/2006/09/15/o...
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