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What are the approximate solutions rounded to the nearest tenth?

What are the approximate solutions rounded to the nearest tenth?-example-1
User MahanTp
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1 Answer

19 votes
19 votes

ANSWER

x = 2.9, y = -1.6 and x = -1.7, y = 6.7

Step-by-step explanation

We are given two equations:


\begin{gathered} 1.8x\text{ + y = 3.6} \\ y=x^2\text{ - 3x - 1} \end{gathered}

From the first equation, we have:

y = 3.6 - 1.8x

Put that in the second, we have:


\begin{gathered} 3.6-1.8x=x^2\text{ - 3x - 1} \\ \text{Collect like terms:} \\ x^2\text{ - 3x - 1 - 3.6 + 1.8x = 0} \\ x^2-\text{ 1.2x - 4.6 = 0} \\ U\sin g\text{ quadratic formula:} \\ x\text{ = }\frac{-b\text{ }\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{where a = 1, b = -1.2, c = -4.6} \\ \Rightarrow\text{ x = }\frac{1.2\text{ }\pm\sqrt[]{(-1.2)^2\text{ - 4(1)(-4.6)}}}{2}\text{ = }\frac{1.2\text{ }\pm\sqrt[]{1.44\text{ + 18.4}}}{2} \\ x\text{ = }\frac{1.2\text{ }\pm\text{ }\sqrt[]{19.84}}{2}\text{ } \\ x\text{ = }\frac{1.2\text{ +4.5}}{2}\text{ and x = }\frac{1.2\text{ - 4.5}}{2} \\ x\text{ = }(5.7)/(2)\text{ and x = }(-3.3)/(2) \\ x\text{ }\cong2.9\text{ and -1.7} \end{gathered}

Put each of those values of x in the first equation:

y = 3.6 - 1.8(2.9) and y = 3.6 - 1.8(-1.7)

y = 3.6 - 5.22 and y = 3.6 + 3.06

y = -1.6 and y = 6.7 (approximately to the nearest tenth)

Therefore, the solutions are:

x = 2.9, -1.6 and x = -1.7, y = 6.7

User Jeffery Ma
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