1. We are given five systems of linear equations, each with three variables. Our goal is to solve each system for the variables. 2. The general approach to solve such systems is to use substitution, elimination, or matrix methods (like Gaussian elimination). Here, we will outline the steps for each system. --- ### System 1: $$\begin{cases} a + b + c = 120 \\ 2a + 3b + c = 250 \\ a + 4b + 2c = 310 \end{cases}$$ 3. Subtract the first equation from the second and third to eliminate $c$: $$ (2a + 3b + c) - (a + b + c) = 250 - 120 \Rightarrow a + 2b = 130 $$ $$ (a + 4b + 2c) - (a + b + c) = 310 - 120 \Rightarrow 3b + c = 190 $$ 4. From the first equation, express $c = 120 - a - b$. 5. Substitute $c$ into $3b + c = 190$: $$ 3b + (120 - a - b) = 190 \Rightarrow 2b - a = 70 $$ 6. Now we have two equations: $$ a + 2b = 130 $$ $$ 2b - a = 70 $$ 7. Add these two equations: $$ (a + 2b) + (2b - a) = 130 + 70 \Rightarrow 4b = 200 \Rightarrow b = 50 $$ 8. Substitute $b=50$ into $a + 2b = 130$: $$ a + 100 = 130 \Rightarrow a = 30 $$ 9. Substitute $a=30$, $b=50$ into $c = 120 - a - b$: $$ c = 120 - 30 - 50 = 40 $$ --- ### System 2: $$\begin{cases} x + 2y + z = 180 \\ 2x + y + 3z = 270 \\ 3x + y + z = 240 \end{cases}$$ 10. From the first equation, express $z = 180 - x - 2y$. 11. Substitute $z$ into the second and third equations: $$ 2x + y + 3(180 - x - 2y) = 270 \Rightarrow 2x + y + 540 - 3x - 6y = 270 \Rightarrow -x - 5y = -270 $$ $$ 3x + y + (180 - x - 2y) = 240 \Rightarrow 3x + y + 180 - x - 2y = 240 \Rightarrow 2x - y = 60 $$ 12. Simplify the first substituted equation: $$ -x - 5y = -270 \Rightarrow x + 5y = 270 $$ 13. Now solve the system: $$ \begin{cases} x + 5y = 270 \\ 2x - y = 60 \end{cases} $$ 14. Multiply the second equation by 5: $$ 10x - 5y = 300 $$ 15. Add to the first equation multiplied by 1: $$ (x + 5y) + (10x - 5y) = 270 + 300 \Rightarrow 11x = 570 \Rightarrow x = \frac{570}{11} = 51.8182 $$ 16. Substitute $x$ into $x + 5y = 270$: $$ 51.8182 + 5y = 270 \Rightarrow 5y = 218.1818 \Rightarrow y = 43.6364 $$ 17. Substitute $x$ and $y$ into $z = 180 - x - 2y$: $$ z = 180 - 51.8182 - 2(43.6364) = 180 - 51.8182 - 87.2728 = 40.909 $$ --- ### System 3: $$\begin{cases} r + s + t = 90 \\ 2r + s + 2t = 160 \\ r + 3s + t = 130 \end{cases}$$ 18. Subtract the first equation from the second and third: $$ (2r + s + 2t) - (r + s + t) = 160 - 90 \Rightarrow r + t = 70 $$ $$ (r + 3s + t) - (r + s + t) = 130 - 90 \Rightarrow 2s = 40 \Rightarrow s = 20 $$ 19. From $r + t = 70$ and $r + s + t = 90$, substitute $s=20$: $$ r + 20 + t = 90 \Rightarrow r + t = 70 $$ 20. This confirms $r + t = 70$. 21. Use the second equation $2r + s + 2t = 160$ with $s=20$: $$ 2r + 20 + 2t = 160 \Rightarrow 2r + 2t = 140 \Rightarrow r + t = 70 $$ 22. We have two variables $r$ and $t$ with one equation $r + t = 70$. Use the third equation $r + 3s + t = 130$ with $s=20$: $$ r + 60 + t = 130 \Rightarrow r + t = 70 $$ 23. All consistent, but infinite solutions along $r + t = 70$. Choose $r=30$, then $t=40$ for example. --- ### System 4: $$\begin{cases} g + f + s = 100 \\ 2g + f + 3s = 220 \\ g + 2f + s = 150 \end{cases}$$ 24. Subtract the first equation from the second and third: $$ (2g + f + 3s) - (g + f + s) = 220 - 100 \Rightarrow g + 2s = 120 $$ $$ (g + 2f + s) - (g + f + s) = 150 - 100 \Rightarrow f = 50 $$ 25. From the first equation, $g + f + s = 100$, substitute $f=50$: $$ g + 50 + s = 100 \Rightarrow g + s = 50 $$ 26. From $g + 2s = 120$ and $g + s = 50$, subtract: $$ (g + 2s) - (g + s) = 120 - 50 \Rightarrow s = 70 $$ 27. Substitute $s=70$ into $g + s = 50$: $$ g + 70 = 50 \Rightarrow g = -20 $$ --- ### System 5: $$\begin{cases} p + c + f = 200 \\ 2p + 3c + f = 400 \\ p + 2c + 2f = 350 \end{cases}$$ 28. Subtract the first equation from the second and third: $$ (2p + 3c + f) - (p + c + f) = 400 - 200 \Rightarrow p + 2c = 200 $$ $$ (p + 2c + 2f) - (p + c + f) = 350 - 200 \Rightarrow c + f = 150 $$ 29. From the first equation, $p + c + f = 200$, and from above $c + f = 150$, substitute: $$ p + 150 = 200 \Rightarrow p = 50 $$ 30. Substitute $p=50$ into $p + 2c = 200$: $$ 50 + 2c = 200 \Rightarrow 2c = 150 \Rightarrow c = 75 $$ 31. Substitute $c=75$ into $c + f = 150$: $$ 75 + f = 150 \Rightarrow f = 75 $$ --- ### Final answers: - System 1: $a=30$, $b=50$, $c=40$ - System 2: $x=\frac{570}{11}$, $y=\frac{240}{11}$, $z=\frac{450}{11}$ (approx. 51.82, 43.64, 40.91) - System 3: $s=20$, $r + t = 70$ (infinite solutions, e.g., $r=30$, $t=40$) - System 4: $g=-20$, $f=50$, $s=70$ - System 5: $p=50$, $c=75$, $f=75$