Math Problem Statement
solve 6-10 using linear system
Solution
Let's solve problems 6 to 10 using linear equations from the image.
Problem 6:
2x + 3y &= 13 \\ 4x - 2y &= 2 \\ \end{aligned}$$ **Solution:** 1. Multiply the first equation by 2 to eliminate $$y$$: $$4x + 6y = 26$$ 2. Subtract the second equation from the modified first equation: $$(4x + 6y) - (4x - 2y) = 26 - 2 \\ 8y = 24$$ 3. Solve for $$y$$: $$y = 3$$ 4. Substitute $$y = 3$$ into the first equation to find $$x$$: $$2x + 3(3) = 13 \\ 2x + 9 = 13 \\ 2x = 4 \\ x = 2$$ **Solution:** $$x = 2$$, $$y = 3$$ ### **Problem 7:** $$\begin{aligned} 2x + 4y &= 10 \\ 4x - y &= -5 \\ \end{aligned}$$ **Solution:** 1. Multiply the second equation by 4 to eliminate $$y$$: $$16x - 4y = -20$$ 2. Add the first equation to the modified second equation: $$(2x + 4y) + (16x - 4y) = 10 + (-20) \\ 18x = -10$$ 3. Solve for $$x$$: $$x = -\frac{10}{18} = -\frac{5}{9}$$ 4. Substitute $$x = -\frac{5}{9}$$ into the first equation to find $$y$$: $$2\left(-\frac{5}{9}\right) + 4y = 10 \\ -\frac{10}{9} + 4y = 10 \\ 4y = 10 + \frac{10}{9} \\ 4y = \frac{100}{9} \\ y = \frac{25}{9}$$ **Solution:** $$x = -\frac{5}{9}$$, $$y = \frac{25}{9}$$ ### **Problem 8:** $$\begin{aligned} 5x + 2y &= 15 \\ 2x - y &= 5 \\ \end{aligned}$$ **Solution:** 1. Multiply the second equation by 2 to eliminate $$y$$: $$4x - 2y = 10$$ 2. Add the first equation to the modified second equation: $$(5x + 2y) + (4x - 2y) = 15 + 10 \\ 9x = 25$$ 3. Solve for $$x$$: $$x = \frac{25}{9}$$ 4. Substitute $$x = \frac{25}{9}$$ into the second equation to find $$y$$: $$2\left(\frac{25}{9}\right) - y = 5 \\ \frac{50}{9} - y = 5 \\ y = \frac{50}{9} - 5 \\ y = \frac{50}{9} - \frac{45}{9} \\ y = \frac{5}{9}$$ **Solution:** $$x = \frac{25}{9}$$, $$y = \frac{5}{9}$$ ### **Problem 9:** $$\begin{aligned} 3x - y &= 7 \\ 2x + 2y &= 6 \\ \end{aligned}$$ **Solution:** 1. Divide the second equation by 2: $$x + y = 3$$ 2. Add this to the first equation to eliminate $$y$$: $$(3x - y) + (x + y) = 7 + 3 \\ 4x = 10$$ 3. Solve for $$x$$: $$x = 2.5$$ 4. Substitute $$x = 2.5$$ into the modified second equation to find $$y$$: $$2.5 + y = 3 \\ y = 0.5$$ **Solution:** $$x = 2.5$$, $$y = 0.5$$ ### **Problem 10:** $$\begin{aligned} x + 2y &= 11 \\ 3x + 4y &= 21 \\ \end{aligned}$$ **Solution:** 1. Multiply the first equation by 3 to eliminate $$x$$: $$3x + 6y = 33$$ 2. Subtract the second equation from the modified first equation: $$(3x + 6y) - (3x + 4y) = 33 - 21 \\ 2y = 12$$ 3. Solve for $$y$$: $$y = 6$$ 4. Substitute $$y = 6$$ into the first equation to find $$x$$: $$x + 2(6) = 11 \\ x + 12 = 11 \\ x = -1$$ **Solution:** $$x = -1$$, $$y = 6$$ Would you like more details or have any other questions? Here are 5 related questions: 1. How do you determine when to add or subtract equations in a system? 2. What are the benefits of the substitution method versus elimination in solving systems? 3. How can you verify the solution of a system of linear equations? 4. What types of systems have no solution or infinite solutions? 5. How do you apply these methods to real-world problems? **Tip:** When solving systems of equations, always simplify the equations as much as possible before applying the elimination or substitution method.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Equations
Systems of Equations
Elimination Method
Substitution Method
Formulas
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Theorems
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Suitable Grade Level
High School
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