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Chapter 13: Problem 17

Solve each system by the substitution method. $$\left\\{\begin{array}{l} x+y=1 \\ (x-1)^{2}+(y+2)^{2}=10 \end{array}\right.$$

Short Answer

Expert verified
The solutions of the system of equations are (4, -3) and (0, 1).

Step by step solution

01

Extract x from the first equation.

The given system of equations is \[\left\{\begin{array}{l} x+y=1, \ (x-1)^{2}+(y+2)^{2}=10. \end{array}\right.\] From the first equation, \(x\) can be isolated as \(x = 1 - y\).
02

Substitute x into the second equation.

Now, replace \(x\) in the second equation with \((1 - y)\) to obtain \[((1 - y)-1)^{2}+(y+2)^{2} = 10\] which simplifies to \[(-y)^{2}+(y+2)^{2} = 10\].
03

Expand and solve the squared equation.

Expand the squared equation to get \[2y^{2}+4y+4 = 10\]. Simplifying this equation gives the quadratic equation \[2y^{2}+4y-6 = 0\]. In order to solve this equation, divide all terms by 2 to get \[y^{2}+2y-3 = 0\]. This can be factored into \[(y + 3)(y - 1) = 0\].
04

Solve for y.

Set each factor equal to zero and solve for \(y\), which gives \(y = -3\) and \(y = 1\).
05

Solve for x using the values of y.

Substitute these solutions back into the first equation \(x = 1 - y\) to solve for \(x\). Substituting \(y = -3\) results in \(x = 4\). Substituting \(y = 1\) results in \(x = 0\).
06

Write the solution as ordered pairs.

The solutions of the system of equations are therefore the ordered pairs \((4, -3)\) and \((0, 1)\).

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Most popular questions from this chapter

If \(f(x)=\frac{1}{3} x-5,\) find \(f^{-1}(x) .\) (Section 8.4, Example 4)

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? $$y=-x^{2}-4 x+4$$

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Find the slope of the line passing through \((-2,-3)\) and \((1,5)\).

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