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own below have a solution? ExI \[ \begin{array}{l}y=5 x+12 \\ 10 x-2 y=1\end{array} \]

Ask by Lewis Guzman. in the United States
Feb 26,2025

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Answer

The system of equations has no solution because the lines are parallel and do not intersect.

Solution

We are given the system of equations:   1) y = 5x + 12   2) 10x – 2y = 1 Step 1. Substitute equation (1) into equation (2):   10x – 2(5x + 12) = 1 Step 2. Distribute the –2:   10x – 10x – 24 = 1 Step 3. Simplify:   0x – 24 = 1 ⟹ –24 = 1 Since –24 ≠ 1, the result is a contradiction, meaning the system has no solution. This indicates that the corresponding lines are parallel and do not intersect.

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The Deep Dive

To determine if the system of equations has a solution, we can use substitution or elimination methods. Starting with the first equation \( y = 5x + 12 \), we can substitute it into the second equation \( 10x - 2y = 1 \). This gives us: \[ 10x - 2(5x + 12) = 1 \] Simplifying this, we get: \[ 10x - 10x - 24 = 1 \] \[ -24 = 1 \] This is a contradiction, which means that the system of equations has no solution. The two lines represented by these equations are parallel and will never intersect! To illustrate further, envision two parallel train tracks: they run alongside each other infinitely but never meet. This means there's no point at which both equations are satisfied simultaneously, confirming our earlier finding of no solution!

Related Questions

ISCELÁNEA cribir, por simple inspección, el resultado de: \( \begin{array}{lll}(x+2)^{2} & \text { 14. }(x+y+1)(x-y-1) & \text { 27. }\left(2 a^{3}-5 b^{4}\right)^{2} \\ (x+2)(x+3) & \text { 15. }(1-a)(a+1) & \text { 28. }\left(a^{3}+12\right)\left(a^{3}-15\right) \\ (x+1)(x-1) & \text { 16. }(m-8)(m+12) & \text { 29. }\left(m^{2}-m+n\right)\left(n+m+m^{2}\right) \\ (x-1)^{2} & \text { 17. }\left(x^{2}-1\right)\left(x^{2}+3\right) & \text { 30. }\left(x^{4}+7\right)\left(x^{4}-11\right) \\ (n+3)(n+5) & \text { 18. }\left(x^{3}+6\right)\left(x^{3}-8\right) & \text { 31. }(11-a b)^{2} \\ (m-3)(m+3) & \text { 19. }\left(5 x^{3}+6 m^{4}\right)^{2} & \text { 32. }\left(x^{2} y^{3}-8\right)\left(x^{2} y^{3}+6\right) \\ (a+b-1)(a+b+1) & \text { 20. }\left(x^{4}-2\right)\left(x^{4}+5\right) & \text { 33. }(a+b)(a-b)\left(a^{2}-b^{2}\right) \\ (1+b)^{3} & \text { 21. }(1-a+b)(b-a-1) & \text { 34. }(x+1)(x-1)\left(x^{2}-2\right) \\ \left(a^{2}+4\right)\left(a^{2}-4\right) & \text { 22. }\left(a^{x}+b^{n}\right)\left(a^{x}-b^{n}\right) & \text { 35. }(a+3)\left(a^{2}+9\right)(a-3) \\ \left(3 a b-5 x^{2}\right)^{2} & \text { 23. }\left(x^{a+1}-8\right)\left(x^{a+1}+9\right) & \text { 36. }(x+5)(x-5)\left(x^{2}+1\right) \\ (a b+3)(3-a b) & \text { 24. }\left(a^{2} b^{2}+c^{2}\right)\left(a^{2} b^{2}-c^{2}\right) & \text { 37. }(a+1)(a-1)(a+2)(a-2) \\ (1-4 a x)^{2} & \text { 25. }(2 a+x)^{3} & \text { 36. }\left(x^{2}-11\right)\left(x^{2}-2\right)\end{array} \) \( \left(a^{2}+8\right)\left(a^{2}-7\right) \)
Algebra Mexico Feb 26, 2025
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