Question

=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 $x-y$ ?

Answer 1

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 y 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 y back into an original equation to solve for x

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 x - y

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! :)

Answer 2

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!