Math Problem Statement
Solution
Let's solve the two systems of equations as requested.
Задание 2
Solve the system using the substitution method:
x - 3 = 3y \\ xy = 6 \end{cases}$$ 1. From the first equation, express $$x$$ in terms of $$y$$: $$x = 3y + 3$$ 2. Substitute $$x = 3y + 3$$ into the second equation: $$(3y + 3)y = 6$$ 3. Expand and simplify: $$3y^2 + 3y = 6$$ $$3y^2 + 3y - 6 = 0$$ $$y^2 + y - 2 = 0$$ 4. Factor the quadratic equation: $$(y + 2)(y - 1) = 0$$ So, $$y = -2$$ or $$y = 1$$. 5. Substitute each value of $$y$$ back into $$x = 3y + 3$$ to find $$x$$: - For $$y = 1$$: $$x = 3(1) + 3 = 6$$ Solution: $$(x, y) = (6, 1)$$ - For $$y = -2$$: $$x = 3(-2) + 3 = -6 + 3 = -3$$ Solution: $$(x, y) = (-3, -2)$$ **Answer for Задание 2**: $$(6, 1)$$ and $$(-3, -2)$$, which matches the given answer. --- ### Задание 3 Solve the system using the method of elimination: $$\begin{cases} 4y^2 - 2xy = 12 \\ y^2 + 8xy = 20 \end{cases}$$ 1. Multiply the second equation by 4 to align the terms: $$4y^2 + 32xy = 80$$ 2. Now we have: $$\begin{cases} 4y^2 - 2xy = 12 \\ 4y^2 + 32xy = 80 \end{cases}$$ 3. Subtract the first equation from the second: $$(4y^2 + 32xy) - (4y^2 - 2xy) = 80 - 12$$ $$34xy = 68$$ $$xy = 2$$ 4. Substitute $$xy = 2$$ back into the first equation: $$4y^2 - 2(2) = 12$$ $$4y^2 - 4 = 12$$ $$4y^2 = 16$$ $$y^2 = 4$$ $$y = \pm 2$$ 5. Substitute each value of $$y$$ into $$xy = 2$$ to find $$x$$: - For $$y = 2$$: $$x \cdot 2 = 2 \Rightarrow x = 1$$ Solution: $$(x, y) = (1, 2)$$ - For $$y = -2$$: $$x \cdot (-2) = 2 \Rightarrow x = -1$$ Solution: $$(x, y) = (-1, -2)$$ **Answer for Задание 3**: $$(1, 2)$$ and $$(-1, -2)$$, which matches the given answer. --- Would you like more details on any step, or do you have any specific questions? --- ### Related Questions 1. How can substitution help solve non-linear systems of equations? 2. What are common methods for solving quadratic equations? 3. How can the elimination method simplify systems with similar terms? 4. What other techniques can be used for systems of equations beyond substitution and elimination? 5. When is factoring effective in solving equations? ### Tip Always double-check each solution by plugging the values back into the original equations to verify accuracy.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Systems of equations
Substitution method
Elimination method
Quadratic equations
Formulas
Substitution formula: x = 3y + 3
Factoring: (y + 2)(y - 1) = 0
Elimination method
Multiplying equations
Theorems
Quadratic equation solution via factoring
Solving non-linear systems using substitution and elimination
Suitable Grade Level
Grades 9-11
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