Solve the above set of equations and choose the correct answer.$$ (I)-5x+4y=3 $$$$ (II)6x-8y=10 $$

$$ x=-4,y=-4\frac{1}{4} $$

$$ x=-4,y=4\frac{1}{4} $$

$$ x=-4\frac{1}{4},y=4 $$

$$ x=-4,y=-4\frac{1}{4} $$

Find the value of x and and band the substitution method.$$ (I)-x+3and=12 $$$$ (II)4x+2and=10 $$

$$ x=\frac{3}{7},y=\frac{29}{7} $$

$$ x=\frac{4}{9},y=\frac{5}{9} $$

Solve the above set of equations and choose the correct answer.$$ (I)\frac{1}{3}x-4y=5 $$$$ (II)x+6y=9 $$

$$ x=11,y=-\frac{1}{3} $$

$$ x=-11,y=-\frac{1}{3} $$

$$ x=-\frac{1}{3},y=-11 $$

$$ x=\frac{1}{3},y=-11 $$

$$ x=11,y=-\frac{1}{3} $$

Find the value of x and and band the substitution method.$$ (I)-4x+4and=15 $$$$ (II)2x+8and=12 $$

$$ x=-\frac{9}{5},y=\frac{39}{20} $$

$$ x=\frac{5}{13},y=-\frac{8}{13} $$

$$ x=-\frac{9}{5},y=\frac{39}{20} $$

Solving with the method of equalization for systems of two linear equations with two unknowns

To solve systems of two linear equations with two unknowns with the equating method, we must arrive at a situation in which one of the coefficients is equal or opposite in the same unknown in the two equations.

How do we do this?

We will arrange the equation in such a way that the types of unknowns lie one on top of the other respectively.
We will carry out a multiplication in one or both equations so that in the two equations there is an identical or opposite coefficient in one of the unknowns (Example of opposite coefficient: $4$ and $-4$ ). The multiplication will be done separately for each term.
We will add or subtract the equations as necessary, in this way we will suppress an unknown and we will obtain an equation with only one unknown (we must pay attention to the signs of subtraction and addition). If the coefficients are equal we will subtract the equations and if they are opposite we will add them.
Let's not forget to place the value found in one of the equations to discover the second unknown.

Detailed explanation

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Test yourself on algebraic solution!

Solve the following equations:

$$ (I)x+y=18 $$

$$ (II)y=13 $$

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Example of the equalization method

$-8y+2x=12$
$8x+2y=-20$
We will see that the coefficients may seem equal or opposite, but they are not in the same unknown, so, in order not to get confused we will arrange the equation so that the types of unknowns are one on top of the other respectively.
$2x-8y=12$
$8x+2y=-20$

We will multiply the first equation by $4$ to arrive at an equal coefficient ( $8$ ) in the unknown $X$ .
We will obtain the following system:
$8x-32y=48$
$8x+2y=-20$
Now, to eliminate the unknown $X$ we can subtract the equations.
Note that if we had caused the coefficients to be opposite, to remove an unknown we would have added the equations.
Let's see that in the free term there is a subtract operation and then another subtraction, which implies more.

We will obtain:
$34y=68$
$ y=2 $
Now let's place what we found in the equation that best suits us and find out the. $X$ :
$2x-8\times 2=12$
$2x=28$
$x=14$
The result is: $y=2 , x=14$

If you are interested in this article you may also be interested in the following articles

Unknowns and Algebraic Expressions
First-degree equations with an unknown variable
What is the unknown of a mathematical equation?
System of Two Linear Equations with Two Unknowns
Linear equation with two unknowns
Graphical solution for a system of linear equations with two unknowns
Algebraic solution for a system of linear equations with two unknowns
Solving with the substitution method for systems of two linear equations with two unknowns
Solving verbal problems with a system of linear equations

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