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Rina8888 [55]
3 years ago
12

=12 \\ 5x-3y=4" align="absmiddle" class="latex-formula">
If (x, \ y) is a solution to the system of equations above, what is the value of x-y ?
Mathematics
2 answers:
Molodets [167]3 years ago
6 0

Answer:

24/25

Step-by-step explanation:

Step 1: Define systems of equation

10x - 16y = 12

5x - 3y = 4

Step 2: Rewrite one of the equations

5x = 4 + 3y

x = 4/5 + 3y/5

Step 3: Solve for <em>y</em> using Substitution

  1. Substitute 2nd rewritten equation into 1: 10(4/5 + 3y/5) - 16y = 12
  2. Distribute the 10 to both terms: 40/5 + 30y/5 - 16y = 12
  3. Simplify the fractions down: 8 + 6y - 16y = 12
  4. Combine like terms (y): 8 - 10y = 12
  5. Subtract 8 on both sides: -10y = 4
  6. Divide both sides by -10: y = 4/-10
  7. Simply the fraction down: y = -2/5

Step 4: Substitute <em>y</em> back into an original equation to solve for <em>x</em>

  1. Substitute: 5x - 3(-2/5) = 4
  2. Multiply: 5x + 6/5 = 4
  3. Subtract 6/5 on both sides: 5x = 14/5
  4. Divide both sides by 5: x = 14/25

Step 5: Check to see if solution set (14/25, -2/5) is a solution.

  1. Substitute into an original equation: 10(14/25) - 16(-2/5) = 12
  2. Multiply each term: 28/5 + 32/5 = 12
  3. Add: 12 = 12

Here, we see that x = 14/25, y = -2/5 and solution (14/25, -2/5) indeed works.

Step 6: Find <em>x</em> <em>- y</em>

x = 14/25

y = -2/5

  1. Substitute: 14/25 - (-2/5)
  2. Simplify (change signs): 14/25 + 2/5
  3. Add: 24/25

Hope this helped! :)

marissa [1.9K]3 years ago
3 0

Answer:

x-y=\frac{24}{25}

Step-by-step explanation:

We have the system:

\left\{        \begin{array}{ll}            10x-16y=12 & \quad \\            5x-3y=4& \quad         \end{array}    \right.

We can solve this system by elimination.

Looking at the system, we can see that the coefficients of the Xs share a LCM.

So, we can multiply the second equation by -2. This yields:

\left\{        \begin{array}{ll}            10x-16y=12 & \quad \\            -2(5x-3y)=-2(4)& \quad         \end{array}    \right.

Multiply:

\left\{        \begin{array}{ll}            10x-16y=12 & \quad \\            -10x+6y=-8 & \quad         \end{array}    \right.

Now, we can add the two equations. This yields:

(10x-10x)+(-16y+6y)=(12+(-8))

Add:

-10y=4

Divide both sides by -12. So, the value of y is:

y=4/-10=-2/5

We can substitute this back into either equation to find x. Let's use the second equation:

5x-3y=4

Substitute -2/5 for y:

5x-3(-\frac{2}{5})=4

Multiply:

5x+\frac{6}{5}=4

Subtract 6/5 from both sides. Note that 4 is the same as 20/5. So:

5x=\frac{20}{5}-\frac{6}{5}

Subtract:

5x=\frac{14}{5}

Multiply both sides by 1/5. So, the value of x is:

x=\frac{14}{25}

Therefore, our solution is:

(\frac{14}{25},- \frac{2}{5})

Where x is 14/25 and y is -2/5.

We want to find:

x-y

Substitute 14/25 for x and -2/5 for y:

=\frac{14}{25}-(-\frac{2}{5})

Simplify. Change 2/5 to 10/25:

=\frac{14}{25}+\frac{10}{25}

Add. So, our answer is:

=\frac{24}{25}

And we're done!

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