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alexandr402 [8]
3 years ago
15

13. Find the solution to the following system of equations using any method: 4x - y = 3 -4x + y = -3

Mathematics
1 answer:
forsale [732]3 years ago
3 0

Start by writing out the system of equations.

  • 4x - y = 3
  • -4x + y = -3

The easiest way to solve this system of equations is the elimination method. This is because both equations contain opposites of each other that you can cancel out.

Add the system of equations together. After adding, everything cancels out. 4x + (-4x) = 0 ; -y + y = 0 ; 3 + (-3) = 0.

You are left with the equation of 0 = 0. Since this is a TRUE statement, this means that there are infinitely many solutions to this system of equations.

If you were left with a false statement, for example 4 = 0, then the answer would be "no solution".

Another way to solve this system of equations is by the substitution method. This method works best if you are given a system of equations with one variable clearly defined.

For example, the system of equations could be:

  • 4x - y = 3
  • y = 3x + 4

y is equal to 3x + 4, so you could easily substitute y into the first equation. The elimination method wouldn't be the best choice because the system of equations are in the standard form Ax + By = C.

Generally any system of equations in the standard form are easier to solve using the elimination method. Hope this helped, good luck!

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LUCKY_DIMON [66]

Answer:

=8\sqrt{15}b^{\frac{7}{2}}

Step-by-step explanation:

\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}

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\sqrt{24b^3}

=\sqrt{24}\sqrt{b^3}

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=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2

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=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

\sqrt{2^3}

=2^{3\cdot \frac{1}{2}

=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2

=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2

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=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2

=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}

2^{\frac{3}{2}+\frac{3}{2}}

=2^3

=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}

b^{\frac{3}{2}+2}

=b^{\frac{7}{2}}

=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}

=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}

=8\sqrt{15}b^{\frac{7}{2}}

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