Leaving Cert Mathematics Algebra: Solve for x: $$\frac{3x + 1}{5}

Solve for x: $$\frac{3x + 1}{5} + \frac{x - 2}{2} = \frac{47}{10}$$ Solve the simultaneous equations: $$x - 5y = -13$$ $$x^2 + y^2 = 13$$ - Leaving Cert Mathematics - Question 4 - 2019

Question 4

$Solve-for-x:--$$\frac{3x-+-1}{5}-+-\frac{x---2}{2}-=-\frac{47}{10}$$--Solve-the-simultaneous-equations:--$$x---5y-=--13$$--$$x^2-+-y^2-=-13$$-Leaving Cert Mathematics-Question 4-2019.png$

Solve for x: $$\frac{3x + 1}{5} + \frac{x - 2}{2} = \frac{47}{10}$$ Solve the simultaneous equations: $$x - 5y = -13$$ $$x^2 + y^2 = 13$$

Worked Solution & Example Answer:Solve for x: $$\frac{3x + 1}{5} + \frac{x - 2}{2} = \frac{47}{10}$$ Solve the simultaneous equations: $$x - 5y = -13$$ $$x^2 + y^2 = 13$$ - Leaving Cert Mathematics - Question 4 - 2019

Step 1

Solve for x:

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Answer

To solve for x, we start with the equation:

$\frac{3x + 1}{5} + \frac{x - 2}{2} = \frac{47}{10}$

Step 1: Find a common denominator.

The common denominator for 5, 2, and 10 is 10. Therefore, we can multiply the entire equation by 10:

$10 \left(\frac{3x + 1}{5}\right) + 10 \left(\frac{x - 2}{2}\right) = 10 \left(\frac{47}{10}\right)$

Step 2: Simplify the equation.

This results in:

$2(3x + 1) + 5(x - 2) = 47$

Step 3: Expand the equation.

Distributing the constants gives us:

$6x + 2 + 5x - 10 = 47$

Step 4: Combine like terms.

To combine the x terms:

$11x - 8 = 47$

Step 5: Isolate x.

Adding 8 to both sides:

$11x = 55$

Divide by 11:

$x = 5$

Step 2

Solve the simultaneous equations:

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Answer

To solve the simultaneous equations:

Identify the equations:
- $x - 5y = -13$ (Equation 1)
- $x^2 + y^2 = 13$ (Equation 2)
Rearrange Equation 1 to express x in terms of y: $x = 5y - 13$
Substitute x in Equation 2:
- Substitute into Equation 2:
$(5y - 13)^2 + y^2 = 13$
Expand and simplify:
- Expanding gives:
$25y^2 - 130y + 169 + y^2 = 13$
- Combine like terms:
$26y^2 - 130y + 156 = 0$
Factoring the quadratic equation:
- We can factor the quadratic as:
$(y - 6)(y - 2) = 0$

This yields:

$y = 6 \text{ or } y = 2$
Substituting back for x:
- For $y = 2$ :
$x = 5(2) - 13 = -3$
- For $y = 6$ :
$x = 5(6) - 13 = 17$
Final Solutions:
- Thus, the solutions are:
$(-3, 2) \text{ and } (17, 6)$

Solve for x: $$\frac{3x + 1}{5} + \frac{x - 2}{2} = \frac{47}{10}$$ Solve the simultaneous equations: $$x - 5y = -13$$ $$x^2 + y^2 = 13$$ - Leaving Cert Mathematics - Question 4 - 2019

Worked Solution & Example Answer:Solve for x: $$\frac{3x + 1}{5} + \frac{x - 2}{2} = \frac{47}{10}$$ Solve the simultaneous equations: $$x - 5y = -13$$ $$x^2 + y^2 = 13$$ - Leaving Cert Mathematics - Question 4 - 2019

Solve for x:

Solve the simultaneous equations:

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