b We will use the Elimination Method to solve this system of equations. It is usually the best choice when one of the variables has equal or opposite coefficients, as they are in the given equation.

{6 x - 2 y = 5 3 x + 2 y = - 2 (I) (II)

SysEqnAdd

$(I):$ Add $(II)$

{6 x - 2 y + (3 x + 2 y) = 5 + (- 2) 3 x + 2 y = - 2

▼

(I):

Solve for

x

RemovePar

$(I):$ Remove parentheses

{6 x - 2 y + 3 x + 2 y = 5 - 2 3 x + 2 y = - 2

AddSubTerms

$(I):$ Add and subtract terms

{9 x = 3 3 x + 2 y = - 2

DivEqn

$(I):$ $LHS / 9 = RHS / 9$

{x = \frac{3}{9} 3 x + 2 y = - 2

ReduceFrac

$(I):$ $\frac{a}{b} = \frac{a / 3}{b / 3}$

{x = \frac{1}{3} 3 x + 2 y = - 2

Having found

x,

we can substitute this into the second equation to find

y .

{x = \frac{1}{3} 3 x + 2 y = - 2 (I) (II)

Substitute

$(II):$ $x = \frac{1}{3}$

{x = \frac{1}{3} 3 (\frac{1}{3}) + 2 y = - 2

▼

(II):

Solve for

y

DenomMultFracToNumber

$(II):$ $3 \cdot \frac{a}{3} = a$

{x = \frac{1}{3} 1 + 2 y = - 2

SubEqn

$(II):$ $LHS - 1 = RHS - 1$

{x = \frac{1}{3} 2 y = - 3

DivEqn

$(II):$ $LHS / 2 = RHS / 2$

{x = \frac{1}{3} y = \frac{- 3}{2}

MoveNegNumToFrac

$(II):$ Put minus sign in front of fraction

{x = \frac{1}{3} y = - \frac{3}{2}

We can check our solution by substituting

x = \frac{1}{3}

and

y = - \frac{3}{2}

into the original system of equations. If the left-hand side and right-hand side are equal in both equations, the solution is correct.

{6 x - 2 y = 5 3 x + 2 y = - 2 (I) (II)

$(I), (II):$ $x = \frac{1}{3}$ , $y = - \frac{3}{2}$

{6 (\frac{1}{3}) - 2 (- \frac{3}{2}) = ? 5 3 (\frac{1}{3}) + 2 (- \frac{3}{2}) = ? - 2

▼

Simplify left-hand side

$(I), (II):$ $2 \cdot \frac{a}{2} = a$

{6 (\frac{1}{3}) - (- 3) = ? 5 3 (\frac{1}{3}) + (- 3) = ? - 2

DenomMultFracToNumber

$(II):$ $3 \cdot \frac{a}{3} = a$

{6 (\frac{1}{3}) - (- 3) = ? 5 1 + (- 3) = ? - 2

Multiply

$(I):$ Multiply

{2 - (- 3) = ? 5 1 + (- 3) = ? - 2

SubNeg

$(I):$ $a - (- b) = a + b$

{2 + 3 = ? 5 1 + (- 3) = ? - 2

AddNeg

$(II):$ $a + (- b) = a - b$

{2 + 3 = ? 5 1 - 3 = ? - 2

AddSubTerms

Add and subtract terms

{5 = 5 - 2 = - 2

Both equations are true, so our solution is correct!

Exercise

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