When solving a system of equations using substitution, there are three steps to follow.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
Substitute this solution into one of the equations and solve for the value of the other variable.

For this exercise,

x

is already isolated in one equation, so we can skip straight to solving!

{4 y = x 3 x - y = 70 (I) (II)

$(II):$ $x = 4 y$

{4 y = x 3 (4 y) - y = 70

$(II):$ Multiply

{4 y = x 12 y - y = 70

$(II):$ Subtract term

{4 y = x 11 y = 70

$(II):$ $LHS / 11 = RHS / 11$

{4 y = x y = \frac{7 0}{1 1}

Now, to find the value of

x,

we need to substitute

y = \frac{7 0}{1 1}

into either one of the equations in the given system. Let's use the first equation.

{4 y = x y = \frac{7 0}{1 1}

$(I):$ $y = \frac{7 0}{1 1}$

{4 (\frac{7 0}{1 1}) = x y = \frac{7 0}{1 1}

$(I):$ Multiply

{\frac{2 8 0}{1 1} = x y = \frac{7 0}{1 1}

$(I):$ Rearrange equation

{x = \frac{2 8 0}{1 1} y = \frac{7 0}{1 1}

The solution to this system of equations is the point

(\frac{2 8 0}{1 1}, \frac{7 0}{1 1}) .

We can also rewrite these solutions as mixed numbers.

{x = \frac{2 8 0}{1 1} y = \frac{7 0}{1 1}

▼

(I):

Write fraction as a mixed number

$(I):$ Write as a sum

{x = \frac{2 7 5 + 5}{1 1} y = \frac{7 0}{1 1}

$(I):$ Split into factors

{x = \frac{2 5 ( 1 1 ) + 5}{1 1} y = \frac{7 0}{1 1}

$(I):$ Write fraction as a mixed number

{x = 25 \frac{5}{1 1} y = \frac{7 0}{1 1}

▼

(II):

Write fraction as a mixed number

$(II):$ Write as a sum

{x = 25 \frac{5}{1 1} y = \frac{6 6 + 4}{1 1}

$(II):$ Split into factors

{x = 25 \frac{5}{1 1} y = \frac{6 ( 1 1 ) + 4}{1 1}

$(II):$ Write fraction as a mixed number

{x = 25 \frac{5}{1 1} y = 6 \frac{4}{1 1}

To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.

{4 y = x 3 x - y = 70 (I) (II)

$(I), (II):$ $x = \frac{2 8 0}{1 1}$ , $y = \frac{7 0}{1 1}$

{4 (\frac{7 0}{1 1}) = ? \frac{2 8 0}{1 1} 3 (\frac{2 8 0}{1 1}) - \frac{7 0}{1 1} = ? 70

$(I), (II):$ Multiply

{\frac{2 8 0}{1 1} = \frac{2 8 0}{1 1} \frac{8 4 0}{1 1} - \frac{7 0}{1 1} = ? 70

$(II):$ Subtract fractions

{\frac{2 8 0}{1 1} = \frac{2 8 0}{1 1} \frac{7 7 0}{1 1} = ? 70

$(II):$ Calculate quotient

{\frac{2 8 0}{1 1} = \frac{2 8 0}{1 1} 70 = 70

Because both equations are true statements, we know that our solution is correct.

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