Subjects algebra

Solve Substitution A401Fd

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1. **Solve the system:** $$\begin{cases} 10x - y = 9 \\ -10x + y = -9 \end{cases}$$ 2. From the first equation, express $y$: $$y = 10x - 9$$ 3. Substitute $y$ into the second equation: $$-10x + (10x - 9) = -9$$ 4. Simplify: $$-10x + 10x - 9 = -9$$ $$0 - 9 = -9$$ $$-9 = -9$$ 5. This is true for all $x$, so the system has infinitely many solutions along the line: $$y = 10x - 9$$ --- 1. **Solve the system:** $$\begin{cases} x + 5y = 6 \\ 9x + 2y = 11 \end{cases}$$ 2. From the first equation, express $y$: $$5y = 6 - x$$ $$y = \frac{6 - x}{5}$$ 3. Substitute $y$ into the second equation: $$9x + 2\left(\frac{6 - x}{5}\right) = 11$$ 4. Multiply both sides by 5 to clear the denominator: $$5 \times 9x + 5 \times 2 \times \frac{6 - x}{5} = 5 \times 11$$ $$45x + 2(6 - x) = 55$$ 5. Distribute: $$45x + 12 - 2x = 55$$ 6. Combine like terms: $$43x + 12 = 55$$ 7. Subtract 12 from both sides: $$43x = 55 - 12$$ $$43x = 43$$ 8. Divide both sides by 43: $$x = \frac{\cancel{43}x}{\cancel{43}} = \frac{43}{43} = 1$$ 9. Substitute $x=1$ back into $y = \frac{6 - x}{5}$: $$y = \frac{6 - 1}{5} = \frac{5}{5} = 1$$ 10. Solution: $$(x,y) = (1,1)$$ --- 1. **Solve the system:** $$\begin{cases} -8x + y = 28 \\ 3x - 4y = -25 \end{cases}$$ 2. From the first equation, express $y$: $$y = 28 + 8x$$ 3. Substitute $y$ into the second equation: $$3x - 4(28 + 8x) = -25$$ 4. Distribute: $$3x - 112 - 32x = -25$$ 5. Combine like terms: $$-29x - 112 = -25$$ 6. Add 112 to both sides: $$-29x = -25 + 112$$ $$-29x = 87$$ 7. Divide both sides by -29: $$x = \frac{87}{\cancel{-29}} \times \frac{-1}{\cancel{-1}} = -\frac{87}{29}$$ 8. Substitute $x = -\frac{87}{29}$ into $y = 28 + 8x$: $$y = 28 + 8 \left(-\frac{87}{29}\right) = 28 - \frac{696}{29}$$ 9. Convert 28 to fraction with denominator 29: $$28 = \frac{812}{29}$$ 10. Calculate $y$: $$y = \frac{812}{29} - \frac{696}{29} = \frac{116}{29}$$ 11. Solution: $$(x,y) = \left(-\frac{87}{29}, \frac{116}{29}\right)$$ --- 1. **Solve the system:** $$\begin{cases} -8x - 3y = 27 \\ -3x + y = 8 \end{cases}$$ 2. From the second equation, express $y$: $$y = 8 + 3x$$ 3. Substitute $y$ into the first equation: $$-8x - 3(8 + 3x) = 27$$ 4. Distribute: $$-8x - 24 - 9x = 27$$ 5. Combine like terms: $$-17x - 24 = 27$$ 6. Add 24 to both sides: $$-17x = 27 + 24$$ $$-17x = 51$$ 7. Divide both sides by -17: $$x = \frac{51}{\cancel{-17}} \times \frac{-1}{\cancel{-1}} = -3$$ 8. Substitute $x = -3$ into $y = 8 + 3x$: $$y = 8 + 3(-3) = 8 - 9 = -1$$ 9. Solution: $$(x,y) = (-3, -1)$$